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ROBINSON'S  MATHEMATICAL  SERIES. 


THE 

PROGRESSIVE 

HIGHER  ARITHMETIC, 

FOR 

SCHOOLS,   ACADEMIES,   AND    MERCANTILE    COLLEGES. 


COMBINING    THB 


ANALYTIC  AND  SYNTHETIC  METHODS; 

AND  rOBMING  A  COMPLETE  TREATISE  OK  ARITHMETICAL 

SCIENCE,     AND    ITS    COMMERCIAL    AND 

BUSINESS  APPLICATIONS. 

BY 

HORATIO   N.  EOBINSON,  LL.  D., 

AUTHOR  OF  WORKS   ON   ALGEBRA,  GEOMETRY  AND  TRIGONOMETRY,   BURYEYINQ  AIT* 
NAVIGATION,  CONIO  SfiCTIONS,  CALCULUS,  ASTRONOMY,  ETC. 


NEW  YORK: 

IVISON,  PUINNEY,  BLA'KEMAN  &  CO^ 

CHICAGO :   a  C.  GRIGGS  ft  00. 

186  7. 


RO  B  I  N  S  O  N'S 


TJie  most  Complete,  most  Practical,  and  most  Scientific  Series  of 
Mathematical  Text-Books  ever  issued  in  this  country, 

►»  ♦  ■» — ■ 

Robinson's  Progressive  Table  Book,  -       -       -      .       . 
Kobinson's  Progressive  Primary  Arithmetic, - 
Robinson's  Progressive  Intellectual  AritLmetic,    - 
Robinson's  Rudiments  of  Written  Arithmetic,      - 
Robinson's  Progressive  Practical  Arithmetic,        -       -       - 

Robinson's  Key  to  Practical  Arithmetic, 

Robinson's  Progressive  Higher  Arithmetic,    -       -       - 

Robinson's  Key  to  Higher  Arithmetic, 

Robinson's  Arithmetical  Examples, 

Robinson's  New   JElementary  Algebra,      -       -       -       .       ^ 

Robinson's  Key  to  Elementary  Algebra, 

Robinson's  University  Algebra, 

Robinson's  Key  to  University  Algebra, 

Robinson's  Wew  University  Algebra, 

Robinson's  JS.ey  to  New  University  Algebra,  •       .       -       - 
Robinson's  New  Geometry  and  Trigonometry,     - 

Robinson's  Surveying  and  Navigation, 

Robinson's  Analyt.  Geometry  and  Conic  Sections, 
Robinson's  DifFeren.  and  Int.  Calculus,  (in  preparations- 
Robinson's  Elementary  Astronomy,   - 

Robinson's  University  Astronomy, 

Robinson's  Mathematical  Operations, 

Robinson's  Key  to  Geometry  and  Trigonometry,   Conic 
Sections  and  Analytical  Geometry, 

Entered,  according  to  Act  of  Congress,  in  the  year  1S60,  by 

DANIEL  W.  riSII   &  J.   11.  FRENCH, 

and  again  in  the  year  1S63,  by 

DANIEL    W.   FI8II,    A.M., 

In  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Nortlicrn 
District  of  the  New  York. 


PREFACE 


This  work  is  intended  to  complete  a  well  graded  and 
progressive  series  of  Arithmetics,  and  to  furnish  to  ad- 
vanced students  a  more  full  and  comprehensive  text-book 
on  the  Science  of  Numbers  than  has  before  been  published ; 
a  work  that  shall  embrace  those  subjects  necessary  to  give 
the  pupil  a  thoroughly  practical  and  scientific  arithmetical 
education,  either  for  the  farm,  the  workshop,  or  a  profes- 
sion, or  for  the  more  difiicult  operations  of  the  counting- 
room  and  of  mercantile  and  commercial  life. 

There  are  two  general  methods  of  presenting  the  ele- 
ments of  arithmetical  science,  the  Sijnthetio  and  the  Ana- 
lytic. Comparison  enters  into  every  operation,  from  the 
simplest  combination  of  numbers  to  the  most  complicated 
problems  in  the  Higher  Mathematics.  Analysis  first 
generalizes  a  subject  and  then  develops  the  particulars  of 
i;vhich  it  consists;  Synthesis  first  presents  particulars, 
from  which,  by  easy  and  progressive  steps,  the  pupil  is  led 
to  a  general  and  comprehensive  view  of  the  subject. 
Analysis  separates  truths  and  properties  into  their  ele- 
ments or  first "  principles ;  Synthesis  constructs  general 
principles  from  particular  cases.  Analysis  appeals  more 
to  the  reason,  and  cultivates  the  desire  to  search  for  first 
principles,  and  to  understarid  the  reason  for  every  process 
rather  than  to  know  the  rule.  Hence,  the  leading  method  in 
an  elementary  course  of  instruction  should  be  the  Synthetic, 
while  in  an  advanced  course  ]t  should  be  the  Analytic. 

The  following  characteristics  of  a  first  class  text-book 
will  be  obvious  to  all  who  examine  this  work:  the  typogra- 

(iii) 


lY  PKEFACE. ' 

pliy  and  mechanical  execxttion ;  the  philosophical  and 
scieyitific  arrangement  of  the  subjects ;  clear  and  concise 
definitions ;  full  and  rigid  analyses ;  exact  and  compre- 
hensive rules;  brief  and  accurate  methods  of  operation: 
the  zuide  range  of  subjects  and  the  large  number  and  prac- 
ticcd  character  of  the  examples — in  a  word,  scientific  ac- 
CUEACY  combined  with  peactical  utility,  throughout  the 
entire  ivork. 

Much  labor  and  attention  have  been  devoted  to  obtain- 
ing correct  and  adequate  information  pertaining  to  mer- 
cantile and  commercial  transactions,  and  the  Government 
Standard  units  of  measures,  weights,  and  money.  The 
counting-room,  the  bank,  the  insurance  and  broker's  office, 
.the  navy  and  ship-yard,  the  manufactory,  the  wharves,  the 
custom-house,  and  the  mint,  have  all  been  visited,  and  the 
most  reliable  statistics  and  the  latest  statutes  have  been 
consulted,  for  the  purpose  of  securing  entire  accuracy  in 
those  parts  of  this  work  which  relate  to  these  subjects 
and  departments.  As  the  result  of  this  thorough  investi- 
gation, many  statements  found  in  most  other  arithmetics 
of  a  similar  grade  will  not  agree  with  the  facts  presented  in 
this  work,  and  simply  because  the  statements  in  these 
other  books  have  been  copied  from  older  works,  while  laws 
and  customs  have  undergone  great  changes  since  the  older 
works  were  written. 

New  material  and  new  methods  will  be  found  in  the  seve- 
ral subjects  throughout  the  entire  work.  Considerable  pro- 
minence has  been  given  to  Percentage  and  its  numerous  ap- 
plications, especially  to  Stocks,  Insurance,  Interest,  Aver- 
aging Accounts,  Domestic  and  Foreign  Exchange,  and  seve- 
ral other  subjects  necessary  to  qualify  students  to  become 
good  accountants  or  commercial  business  men.  And  while 
this  work  may  embrace  many  subjects  not  necessary  to  the 


PREFACE.  y 

course  usually  prescribed  in  Mercantile  and  Commercial 
Colleges,  yet  those  subjects  requisite  to  make  good  account- 
ants, and  which  have  been  taught  orally  in  that  class  of 
institutions  from  want  of  a  suitable  text-book,  are  fully  dis- 
cussed and  practically  applied  in  this  work ;  and  it  is  there- 
fore believed  to  be  better  adapted  to  the  wants  of  Mercan- 
tile Colleges  than  any  similar  w^ork  yet  published.  And 
while  it  is  due,  it  is  also  proper  here  to  state  that  J.  C. 
Porter,  A.  M.,  an  experienced  and  successful  teacher  of 
Mathematics  in  this  State,  and  formerly  professor  of  Com- 
mercial Arithmetic,  in  Iron  City  Commercial  College,  Pitts- 
burgh, Penn.,  has  rendered  valuable  aid  in  the  preparation 
of  the  above-named  subjects,  and  of  other  portions  of  the 
work.  He  is  likewise  the  author  of  the  Factor  Table  on 
pages  72  and  73,  and  of  the  new  and  valuable  improvement 
in  the  method  of  Cube  Eoot. 

Teachers  entertain  various  views  relative  to  having  the 
answers  to  problems  and  examples  inserted  in  a  text-book. 
Some  desire  the  answers  placed  immediately  after  the  ex- 
amples ;  others  wish  them  placed  together  in  the  back  part 
of  the  book;  and  still  others  desire  them  omitted  alto- 
gether. All  these  methods  have  their  advantages  and  their 
disadvantages. 

If  all  the  answers  are  given,  there  is  danger  that  the 
pupil  will  become  careless,  and  not  depend  enough  upon  the 
accuracy  of  his  own  computations.  Hence  he  is  liable  to 
neglect  the  cultivation  of  those  habits  of  patient  investiga- 
tion and  self-reliance  which  would  result  from  his  being 
obliged  to  test  the  truth  and  accuracy  of  his  own  processes 
by  proof, — the  only  test  he  will  have  to  depend  upon  in  all 
the  computations  in  real  business  transactions  in  after  life. 
Besides,  the  work  of  proving  the  correctness  of  a  result  is 
often  of  quite  as  much  value  to  the  pupil  as  the  work  of 
1* 


yi  PREFACE. 

performing  the  operation ;  as  the  proof  may  render  simple 
and  clear  some  part  or  the  whole  of  an  operation  that  was 
before  complicated  and  obscure. 

If  answers  ara  placed  in  the  back  part  of  the  book,  the 
pupil  will  at  once  refer  to  them  whenever  he  is  in  any  doubt 
or  difficulty  in  performing  an  operation.  Hence  the  object 
aimed  at  is  not  accomplished  by  placing  the  answers  to- 
gether in  this  manner. 

Again,  if  all  the  answers  are  omitted,  the  pupil  may  be- 
come involved  in  doubt  and  uncertainty,  and  acquire  a 
distaste  for  the  study ;  and  from  this  discouragement,  sub- 
sequently make  but  limited  advancement  in  Mathematical 
Science. 

In  order,  therefore,  that  pupils  may  receive  the  advan- 
tages of  both  methods,  the  answers  to  nearly  one  half 
of  the  examples  in  this  book  are  omitted.  They  will  be 
found,  together  with  full  and  clear  solutions  of  all  the 
examples,  in  a  Key  to  this  work,  which  has  been  prepared 
for  the  use  of  teachers  and  private  learners. 

Many  valuable  hints  and  suggestions  which  have  been 
received  from  teachers  and  friends  of  education,  have 
been  incorporated  into  this  work.  The  author  desires  to 
make  especial  acknowledgment  of  the  valuable  services 
rendered  in  the  preparation  of  this  work  by  D.  W.  Fish,  A.M., 
of  Rochester,  N.  Y.,  a  gentleman  who  has  had  long  and 
successful  experience  as  a  teacher,  and  an  intimate  ac- 
quaintance with  the  plans  and  operations  of  some  of  the 
best  schools  in  the  country. 

Augii-st  1,   1860. 


CONTENTS, 


1^- 
•  PAG> 

Definitions 11 

Signs 13 

Axioms , 14 

Notation  and  Numeration ,. , 15 

SIMPLE    NUMBERS. 

Addition 23 

Adding  two  or  more  celumns  at  one  operation 27 

Subtraction 30 

Two  or  more  subtrahends , 33 

Multiplication 33 

Powers  of  Numbers 89 

Continued  Multiplication.. , 40 

Contractions  in  Multiplication 41 

Division 47 

Abbreviated  Long  Division 60 

Successive  Division 55 

Contractions  in  Division 55 

General  Problems  in  Simple  Numbers , 61 

PROPERTIES    OF   NUMBERS. 

Exact  Divisors 65 

Prime  Numbers 6S 

Table  of  Prime  Numbers 70 

Factoring 70 

Factor  Table 72 

Greatest  Common  Divisor 76 

Least  Common  Multiple 82 

Cancellation 86 

FRACTIONS. 

Definitions,  Notation  and  Numeration 89 

Reduction 92 

Addition 99 

(yii) 


Viii  CONTENTS. 

PAQB 

Subtraction 101 

Theory  of  Multiplication  and  Division 103 

Multiplication 104 

Division 107 

Greatest  Common  Divisor Ill 

Least  Common  Multiple 112 

DECIMALS. 

Notation  and  Numeration 117 

Reduction , 121 

Addition 124 

Subtraction 126 

Multiplication 127 

Contracted  Multiplication 128 

Division 132 

Contracted  Division 134 

Circulating  Decimals 136 

Reduction  of  Circulating  Decimals 139 

Addition  and  Subtraction  of  Circulating  Decimals 142 

Multiplication  and  Division  of  Circulating  Decimals 144 

UNITED    STATES    MONEY. 

Notation  and  Numeration 145 

Reduction 147 

Operations 147 

Problems 150 

Ledger  Accounts 153 

Accounts  and  Bills 153 

Continued  Fractions 161 

COMPOUND    NUMBERS. 

Measures  of  Extension 164 

Measures  of  Capacity 170 

Measures  of  Weight 171 

Measure  of  Time 175 

Measure  of  Angles 177 

Miscellaneous  Tables 178 

Government  Standards  of  Measures  and  Weights 179 

English  Measures  and  Weights 182 

French  Measures  and  Weights 184 

Money  and  Currencies 187 

Reduction 192 

Reduction  Descending 192 


CONTENTS,  Jx 

PAGE 

Reduction  Ascending , 199 

Addition 206 

Subtraction 209 

Multiplication 214 

Division ,.  216 

Longitude  and  Time 218 

DUODECIMALS. 

Addition  and  Subtraction 22'?' 

Multiplication 223 

Division 230 

SHORT    METHODS. 

For  Subtraction 232 

For  Multiplication 233 

For  Division 241 

RATIO.  '  243 

PROPORTION.  24r 

Cause  and  Effect 249 

Simple  Proportion 249 

Compound  Proportion 253 

PERCENTAGE. 

Notation 259 

General  Problems 260 

Applications 268 

Commission 2G8 

Stocks 272 

Stock-jobbing 273 

Instalments,  Assessments,  and  Dividends 276 

Stock  Investments 279 

Gold  Investments 2S5 

Profit  and  Loss 287 

Insurance 291 

Life  Insurance 293 

Life  Table 295 

Endowment  Assurance  Table 206 

Taxes 29S 

General  Average 801 

Custom  House  Business 803 

Simple  Interest 807 


X 


CONTENTS. 


PAGa 

Partial  Payments  or  Indorsements 314 

Savings  Banks  Accounts 319 

Compound  Interest 321 

Compound  Interest  Table «... 823 

Problems  in  Interest 324 

Discount 328 

Banking 330 

ExchuKge « -•.. S37 

Direct  Exchange. 339 

Table  of  Foreign  Corns  and  Money ,..^  342 

Arbitrated  Exchange , 348 

Equation  of  Payments : 352 

Simple  Equations 352 

Compound  Equations 857 

Account  Sales 861 

Settlement  of  Accounts  Current 363 

Partnership. c , 864 


ALLIGATION. 


870 


INVOLUTION. 


879 


EVOLUTION.  881 

Square  Root 882 

Contracted  Method 386 

Cube  Root 887 

Contracted  Method , 892 

Applications  of  the  Square  and  Cube  Roots 898 

SERIES.  406 

Arithmetical  Progression 408 

Geometrical  Progression 411 

Compound  Interest  by  Geometrical  Progression 415 

Annuities 416 

Annuities  at  Simple  Interest. 417 

Annuities  at  Compound  Interest 421 

Miscellaneous  Examples 422 

Metric  System 429 


HIGHER  ARITHMETIC. 


DEFINITIONS. 

I,  Quantity  is  any  thing  that  can  be  increased,  diminished,  or 
measured ;  as  distance,  space,  weight,  motion,  time. 

3,    A  Unit  is  one,  a  single  thing,  or  a  definite  quantity. 

3.   A  Number  is  a  unit,  or  a  collection  of  units. 

-4.  The  Unit  of  a  Number  is  one  of  the  collection  constituting 
the  number.     Thus,  the  unit  of  34  is  1 ;  of  34  days  is  1  day. 

5.  An  Abstract  Number  is  a  number  used  without  reference 
to  any  particular  thing  or  quantity;  as  3,  24,  756. 

©•  A  Concrete  Number  is  a  number  used  with  reference  to 
some  particular  thing  or  quantity;  as  21  hours,  4  cents,  230  miles. 

7.  Unity  is  the  unit  of  an  abstract  number. 

8.  The  Denomination  is  the  name  of  the  unit  of  a  concrete 
number. 

O.  A  Simple  Number  is  either  an  abstract  number,  or  a  con- 
crete number  of  but  one  denomination;  as  48,  52  pounds,  36  days. 

10.  A  Compound  Number  is  a  concrete  number  expressed  in 
two  or  more  denominations ;  as,  4  bushels  3  pecks,  8  rods  4  yards 
2  feet  3  inches. 

II.  An  Integral  Number,  or  Integer,  is  a  number  which  ex- 
presses whole  things;  as  5,  12  dollars,  17  men. 

12.  A  Fractional  Number,  or  Fraction,  is  a  number  which 
expresses  equal  parts  of  a  whole  thing  or  quantity;  as  J,  f  of  a 
pound,  ~^^  of  a  bushel. 

13.  Like  Numbers  have  the  same  kind  of  unit,  or  express  the 
same  kind  of  quantity.  Thus,  74  and  16  are  like  numbers;  so 
are  74  pounds,  16  pounds,  and  12  pounds;  also,  4  weeks  3  days,  and 
16  minutes  20  seconds,  both  being  used  to  express  units  of  time. 

14.  Unlike  Numbers  have  different  kinds  of  units,  or  are  used 

(11) 


12  SIMPLE  NUMBERS. 

to  express  different  kinds  of  quantity.  Thus,  36  miles,  and  15 
days ;  5  hours  36  minutes,  and  7  bushels  3  pecks. 

lo.  A  Power  is  the  product  arising  from  multiplying  a  number 
by  itself,  or  repeating  it  any  number  of  times  as  a  factor. 

10.   A  Boot  is  a  factor  repeated  to  produce  a  power. 

17.  A  Scale  is  the  order  of  progression  on  which  any  system 
of  notation  is  founded.     Scales  are  uniform  and  varyino-. 

18.  A  Uniform  Scale  is  one  in  which  the  order  of  progression 
is  the  same  throughout  the  entire  succession  of  units. 

19.  A  Varying  Scale  is  one  in  which  the  order  of  progression 
is  not  the  same  throughout  the  entire  succession  of  units.  ° 

20.  A  Decimal  Scale  is  one  in  which  the  order  of  progression 
is  uniformly  ten. 

21.  Mathematics  ia  the  science  of  quantity. 

The  two  fundamental  branches  of  Mathematics  are  Geometry 
and  Arithmetic.  Geometry  considers  quantity  with  reference  to 
positions,  form,  and  extension.  Arithmetic  considers  quantity  as 
an  assemblage  of  definite  portions,  and  treats  only  of  those  condi- 
tions and  attributes  which  may  be  investigated  and  expressed  by 
numbers.     Hence, 

22.  Arithmetic  is  the  Science  of  numbers,  and  the  Art  of 

computation. 

KoTE  yi,  Arithmetic  treats  of  operations  on  abstract  numbers  it  is  a  sci- 
ence,  and  is  then  called  Pure  Arithmetic.  When  it  treats  of  operations  on  con- 
Crete  numbers  it  ,s  an  art,  «nd  ig  then  cnWed  Applied  ArithLtic.  Pure  and 
Applied  Arithmetic  are  also  called  Theoretical  and  Practical  Arithmetic. 

23.  A  Demonstration  is  a  process  of  reasoning  by  which  a 
truth  or  principle  is  established. 

24.  An  Operation  is  a  process  in  which  figures  are  employed 
to  make  a  computation,  or  obtain  some  arithmetical  result. 

25.  A  Problem  is  a  question  requiring  an  operation. 

20.    A  Rule  is  a  prescribed  method  of  performing  an  operation. 

27.   Analysis,  in  arithmetic,  is  the  process  of  investigating 

principles,  and  solving  problems,  independently  of  set  rules. 

^  28.    The  Five  Fundamental  Operations  of  Arithmetic  are, 

Notation  and  Numeration,  Addition,  Subtraction,  Multiplication, 

and  Division. 


DEFINITIONS. 


SIGNS. 


13 


29.  A  Sign  is  a  character  indicating  the  relation  of  numbers, 
or  an  operation  to  be  performed. 

30.  The  Sign  of  Numeration  is  the  comma  (,).  It  indicates 
that  the  figures  set  off  by  it  express  units  of  the  same  general  name, 
and  are  to  be  read  together,  as  thousands^  millionsy  hillions,  etc. 

31.  The  Decimal  Sign  is  the  period  (.).  It  indicates  that 
the  number  after  it  is  a  decimal. 

33.  The  Sign  of  Addition  is  the  perpendicular  cross,  +,  called 
plus.  It  indicates  that  the  numbers  connected  by  it  are  to  be 
added )  as  3  +  5  +  7,  read  3  plus  5  plus  7. 

33.  The  Sign  of  Subtraction  is  a  short  horizontal  line,  — , 
called  minus.  It  indicates  that  the  number  after  it  is  to  be  sub- 
tracted from  the  number  before  it;  as  12  —  7,  read  12' minus  7. 

34.  The  Sign  of  Multiplication  is  the  oblique  cross,  x  .  It 
indicates  that  the  numbers  connected  by  it  are  to  be  multiplied 
together;  as  5  X  3  x  9,  read  5  multiplied  by  3  multiplied  by  9. 

33.  The  Sign  of  Division  is  a  short  horizontal  line,  with  a 
point  above  and  one  below,  -^-,  It  indicates  that  the  number 
before  it  is  to  be  divided  by  the  number  after  it;  as  18  -f-  6,  read 
18  divided  by  6. 

Division  is  also  expressed  by  writing  the  dividend  ahove,  and 
the  divisor  helow,  a  short  horizontal  line.  Thus,  ^g®,  read  18 
divided  by  6. 

36.  The  Sign  of  Eq[uality  is  two  short,  parallel,  horizontal 
lines,  =.  It  indicates  that  the  numbers,  or  combinations  of 
numbers,  connected  by  it  are  equal;  as  4  +  8  =  15  —  3,  read  4 
plus  8  is  equal  to  15  minus  3.  Expressions  connected  by  the 
sign  of  equality  are  called  equations. 

37.  The  Sign  of  Aggregation  is  a  parenthesis,  (  ).  It  indi- 
cates that  the  numbers  included  within  it  are  to  be  considered 
together,  and  subjected  to  the  same  operation.  Thus,  (8  +  4)  x  5 
indicates  that  both  8  and  4,  or  their  sum,  is  to  be  multiplied  by  5. 

A  vinculum  or  bar,  ,  has  the  same  signification.     Thus, 

7x9-T-3  =  21. 

2 


14  SIMPLE  NUMBERS. 

38.  The  Sign  of  Ratio  is  two  points,  :  .  Thus,  7  :  4  is  read, 
the  ratio  of  7  to  4. 

39.  The  Sign  of  Proportion  is  four  points,  :  :  .  Thus, 
3  :  G  :  :  4  :  8,  is  read,  3  is  to  6  as  4  is  to  8. 

40.  The  Sign  of  Involution  is  a  number  written  above,  and  a 
little  to  the  right,  of  another  number.  It  indicates  the  power  to 
which  the  latter  is  to  be  raised.  Thus,  12^  indicates  that  12  is 
to  be  taken  3  times  as  a  factor;  the  expression  is  equivalent  to 
12  X  12  X  12.  The  number  expressing  the  sign  of  involution  is 
called  the  Index  or  Exponent. 

41.  The  Sign  of  Evolution,  v/,  is  a  modification  of  the  letter  r. 
It  indicates  that  some  root  of  the  number  after  it  is  to  be  extracted. 
Thus,  v/25  indicates  that  the  square  root  of  25  is  to  be  extracted; 
-^64  indicates  that  the  cube  root  of  64  is  to  be  extracted. 

AXIOMS. 

42.  An  Axiom  is  a  self-evident  truth.  The  principal  axioms 
required  in  arithmetical  investigations  are  the  following : 

1.  If  the  same  quantity  or  equal  quantities  be  added  to  equal 
quantities,  the  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted  from 
equal  quantities,  the  remainders  will  be  equal. 

8.  If  equal  quantities  be  multiplied  by  the  same  number,  the 
products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same  number,  the  quo- 
tients will  be  equal. 

5.  If  the  same  number  be  added  to  a  quantity  and  subtracted 
from  the  sum,  the  remainder  will  be  that  quantity. 

6.  If  a  quantity  be  multiplied  by  a  number  and  the  product 
divided  by  the  same  number,  the  quotient  will  be  that  quantity. 

7.  Quantities  which  are  respectively  equal  to  any  other  quantity 
are  equal  to  each  other. 

8.  Like  powers  or  like  roots  of  equal  quantities  are  equal. 

9    The  whole  of  any  quantity  is  greater  than  any  of  its  parts. 
10.  The  whole  of  any  quantity  is  equal  to  the  sum  of  all  its 
parts. 


NOTATION  AND  NUMERATION.  ^5 

NOTATION  AND  NUMERATION. 

43.  Notation  is  p.  system  of  writing  or  expressing  numbers  by 
characters ;  and, 

4:4,  Numeration  is  a  method  of  reading  numbers  expressed 
by  characters. 

45.  Two  systems  of  notation  are  in  general  use  —  the  Roman 
and  the  Arabic, 

Note.  —  The  Romnn  Notntion  is  supposed  to  have  been  first  used  hy  the 
Roirifins  ;  hence  its  name.  The  Arabic  Notation  was  first  introduced  into  Europe 
by  the  Moors  or  Arabs,  who  conquered  and  held  possession  of  Spain  during  tlio 
llih  century.  It  received  the  attention  of  scientific  men  in  Italy  at  the  bepin- 
nit)g  of  the  l.'^th  century,  and  was  soon  afterward  adopted  in  most  European 
countries.  Formerly  it  was  supposed  to  be  an  invention  of  the  Arabs;  but 
investicrations  have  shown  that  the  Arabs  adopted  it  from  the  Hindoos,  among 
whom  it  has  been  in  use  more  than  2000  years.  From  this  undoubted  origin  it 
is  eometimes  called  the  Indian  Notation. 

THE   ROMAN   NOTATION. 

4:6.  Employs  seven  capital  letters  to  express  numbers.  Thus, 
Letters,         I  Y  X  L  C  D  M 

Values,  one,  five,  ten,  fifty,       ,^Xd,    hunrred,    thornd. 

417.  The  Roman  notation  is  founded  upon  five  principles,  as 
follows : 

1st.  Eepeatino^  a  letter  repeats  its  value.  Thus,  II  represents 
two,  XX  twenty,  CCC  three  hundred. 

2d.  If  a  letter  of  any  value  be  placed  a/fer  one  of  greater  value, 
its  value  is  to  be  united  to  that  of  the  greater.  Thus,  XI  repre- 
sents eleven,  LX  sixty,  DC  six  hundred. 

3d.  If  a  letter  of  any  value  be  placed  he/ore  one  of  greater  value, 
its  value  is  to  be  taJcen  from  that  of  the  greater.  Thus,  IX  repre- 
sents nine,  XL  forty,  CD  four  hundred. 

4th.  If  a  letter  of  any  value  be  placed  between  two  letters,  each 
of  greater  value,  its  value  is  to  be  taken  from  the  united  value  of 
the  other  two.  Thus,  XI Y  represents  fourteen,  XXIX  twenty- 
nine,  XCIY  ninety-four. 

5th.  A  bar  or  dash  placed  over  a  letter  increases  its  value  one 
thousand  fold.  Thus,  Y  signifies  five,  and  Y  five  thousand;  L 
fifty,  and  L  fifty  thousand. 


16 


SIMPLE  NUMBEES. 


TABLE 

or    ROMAN    NOTATION. 

I  is 

One. 

XX  is 

i  Twenty. 

II  '' 

Two. 

XXI  ' 

'  Twenty-one. 

III  - 

Three. 

XXX  * 

=  Thirty. 

IV  " 

Four. 

XL  * 

*  Forty. 

V  - 

Five. 

L  * 

'  Fifty. 

VI  " 

Six. 

LX  ' 

*  Sixty. 

VII  " 

Seven. 

LXX  * 

*  Seventy. 

VIII  " 

Eight. 

LXXX 

=*  Eighty. 

IX.  ^' 

Nine. 

XC 

"  Ninety. 

X  - 

Ten. 

C 

"  One  hundred. 

XI  - 

Eleven. 

cc 

"  Two  hundred. 

XII  " 

Twelve. 

D 

"  Five  hundred. 

XIII  - 

Thirteen. 

DC 

"  Six  hundred. 

XIV  " 

Fourteen. 

M 

"  One  thousand.       [dred. 

XV  '' 

Fifteen. 

MC 

"  One  thousand  one  hun- 

XVI  '' 

•   Sixteen. 

MM 

**  Two  thousand. 

XVII  - 

•  Seventeen. 

X 

"  Ten  thousand. 

s:viii  ' 

'  Eighteen. 

C 

"  One  hundred  thousand. 

XIX  '' 

Nineteen. 

M 

"  One  million. 

KoTRS. — 1.  Though  the  letters  used  in  the  above  table  have  been  employed 
as  the  Roman  numerals  for  man^'  centuries,  the  marks  or  characters  used  origi- 
nally in  this  notation  are  as  follows : 

Modern  numerals,  I         V         X         L         C         D         M 

Primitive  characters,  |  V  X  L         C         N         M 

2.  The  system  of  Roman  Notation  is  not  well  adapted  to  the  purposes  of  nu- 
merical calculation;  it  is  principally  confined  to  the  numbering  of  chapters  and 
sections  of  books,  public  documents,  etc. 


EXAMPLES   FOR   PRACTICE. 

Express  the  following  numbers  by  the  Roman  notation: 

1.  Fourteen.  6.  Fifty-one. 

2.  Nineteen.  7.  Eighty-eight. 
8.  Twenty-four.  8.  Seventy-three. 
4.  Thirty-nine.  9.  Ninety-five. 

6.  Forty-six.  10.  One  hundred  one. 

11.  Five  hundred  fifty-five. 

12.  Seven  hundred  ninety-eight. 

13.  One  thousand  three. 

14.  Twenty  thousand  eight  hundred  forty-five. 


NOTATION  AND  NUMERATION.  j'j 

THE   ARABIC    NOTATION 

48.  Employs  ten  characters  or  figures  to  express  numbers. 
Thus, 

Figures,  0        123456789 

,  ^     naught      one,      two,     three,    four,     five,      six,     seven,    eight,    nine, 
values.      \     ^^^^^^ 

49.  The  cipher,  or  fir^t  character,  is  called  naught,  because  it 
has  no  value  of  its  own.  It  is  otherwise  termed  nothing,  and  zero. 
The  other  nine  characters  are  called  significant  figures,  because 
each  has  a  value  of  its  own.  They  are  also  called  digits,  a  word 
derived  from  the  Latin  term  digitus,  which  signifies  finger. 

50.  The  ten  Arabic  characters  are  the  Alphabet  of  Arithmetic. 
Used  independently,  they  can  express  only  the  nine  numbers  that 
correspond  to  the  names  of  the  nine  digits.  But  when  combined 
according  to  certain  principles,  they  serve  to  express  all  numbers. 

51.  The  notation  of  all  numbers  by  the  ten  figures  is  accom- 
plished by  the  formation  of  a  series  of  units  of  different  values,  to 
which  the  digits  may  be  successively  applied.  First  ten  simple 
units  are  considered  together,  and  treated  as  a  single  superior 
unit;  then,  a  collection  often  of  these  new  units  is  taken  as  a  still 
higher  unit;  and  so  on,  indefinitely.  A  regular  series  of  units,  in 
ascending  orders,  is  thus  formed,  as  shown  in  the  following 

TABLE    OF    UNITS. 

Primary  units  are  called  units  of  the- first    order. 

Ten  units  of  the  first     order  make  1  unit    "     "    second    ** 
Ten     "      "     "    second     ''         *'      1     ''      "     "    third       " 
Ten     "      "     "    third        "         "      1     "      "     "    fourth     " 
etc.,        etc.  etc.,        etc. 

59.  The  various  orders  of  units,  when  expressed  by  figures, 
are  distinguished  from  each  other  by  their  location,  or  the  place 
they  occupy  in  a  horizontal  row  of  figures.  Units  of  the  first  order 
are  written  at  the  right  hand ;  units  of  the  second  order  occupy 
the  second  place;  units  of  the  third  order  the  third  place;  and  so 
on,  counting  from  right  to  left,  as  shown  on  the  following  page : 


J8  SIMPLE  NUMBERS. 


t 


•i^       -o       ^       "B 


S        S        ^ 


000000000 

53.  In  this  notation  we  observe  — 

1st.  That  a  figure  written  in  the  place  of  any  order,  expresses 
as  many  units  of  that  order  as  is  denoted  by  the  name  of  the  figure 
used.  Thus,  436  expresses  4  units  of  the  3d  order,  3  units  of  the 
2d  order,  and  6  units  of  the  1st  order. 

2d.  The  cipher,  having  no  value  of  its  own,  is  used  to  fill  the 
places  of  vacant  orders,  and  thus  preserve  the  relative  positions  of 
the  significant  figures.  Thus,  in  50,  the  cipher  shows  the  absence 
of  simple  units,  and  at  the  same  time  gives  to  the  figure  5  the 
local  value  of  the  second  order  of  units. 

54,  Since  the  number  expressed  by  any  figure  depends  upon 
the  place  it  occupies,  it  follows  that  figures  have  two  values, 
Simple  and  Local. 

55.  The  Simple  Value  of  a  figure  is  its  value  when  taken 
alone ;  thus,  4,  7,  2. 

56,  The  Local  Value  of  a  figure  is  its  value  when  used  with 

another  figure  or  figures  in  the  same  number.     Thus,  in  325,  the 

local  value  of  the  3  is  300,  of  the  2  is  20,  and  of  the  5  is  5  units. 

Note. — When  a  figure  occupies  units'  place,  its  simple  and  local  values  are 
the  same. 

57*  The  leading  principles  upon  which  the  Arabic  notation 
is  founded  are  embraced  in  the  following 

GENERAL   LAWS. 

I.  All  numbers  are  expressed  hy  applying  the  ten  figures  to  dif- 
ferent orders  of  units. 

II.  The  different  orders  of  units  increase  from  right  to  left,  and 
decrease  from  left  to  right,  in  a  tenfold  ratio. 

III.  Every  removal  of  a  figure  one  2)lace  to  the  left,  increases  its 
local  value  tenfold;  and  every  removal  of  a  figure  one  place  to  the 
right,  diminishes  its  local  value  tenfold. 


NOTATION  AND  NUMERATION.  19 

08.    In  numerating,  or  expressing  numbers  verbally,  the  various 
orders  of  units  have  the  following  names : 

ORDERS.  NAMES. 

1st  order  is  called  Units, 

2d    order  "      "  Tens. 

3d    order  "      "  Hundreds.' 


4th  order  "  **  Thousands.^ 

5th  order  "  "  Tens  of  thousands. 

6th  order  "  "  Hundreds  of  thousands. 

7th  order  "  "  Millions.^ 

8th  order  "  "  Tens  of  millions. 

9th  order  "  **  Hundreds  of  millions, 

etc.,  etc.  etc.,     etc. 


1} 


59.  This  method  of  numerating,  or  naming,  groups  the  suc- 
cessive orders  into^eriWs  of  three  figures  each,  there  being  three 
orders  of  thousands,  three  orders  of  millions,  and  so  on  in  all 
higher  orders.  These  periods  are  commonly  separated  by  commas, 
as  in  the  following  table,  which  gives  the  names  of  the  orders 
and  periods  to  the  twenty-seventh  place. 


c3 


0 


d  "^  '^  .2  ;5 

cr"  icr"  -J5  pO  a  :3  0 


98, 7  65, 4  32, 109, 876, 556, 789, 012, 3  45 

ninth    eighth     seventh      sixth        fifth       fourth       third       pecond        first 
period,  period,      period,     period,    period,     period,      period,     period.      period. 

NoTR.  —  This  is  the  French  method  of  numerating,  and  is  the  one  in  pfeneral 
use  in  this  country.     The  English  numerate  by  periods  of  six  figures  each. 

60.  The  names  of  the  periods  are  derived  from  the  Latin 
numerals.  The  twenty-two  given  on  the  following  page  extend 
tlie  numeration  table  to  the  sixty-sixth  place  or  order,  inclusive. 


20 


SIMPLE  NUMBERS. 


PERIODS. 

NAMES. 

PERIODS. 

NAMES. 

1st 

Units. 

12th 

Decillions. 

2d 

Thousands. 

13th 

Undecillions. 

3d 

Millions. 

14th 

Duodecillions. 

4th 

Billions. 

15th 

Tredecillions. 

5th 

Trillions. 

16th 

Quatuordecillions 

6th 

Quadrillions. 

17th 

Quindecillions. 

7th 

Quintillions. 

18th 

Sexdecillions. 

8th 

Sextillions. 

19th 

Septendecillions. 

9th 

Septillions. 

20th 

Octodecillions. 

10th 

Octillions. 

21st 

Novendecillions. 

11th 

Nonillions. 

22d 

Vigintillions. 

61.  From  this  analysis  of  the  principles  of  Notation  and  Nume- 
ration, we  derive  the  following  rules : 

RULE   FOR   NOTATION. 

I.  Beginning  at  the  left  hand,  write  the  figures  belonging  to  the 
highest  period. 

II.  Write  the  hundreds^  tenSy  and  units  of  each  successive  period 
in  their  order,  placing  a  cipher  wherever  an  order  of  units  is 
omitted. 

RULE    FOR    NUMERATION. 

I.  Separate  the  number  into  periods  of  three  figures  each,  com^ 
mencing  at  the  right  hand. 

II.  Beginning  at  the  left  hand,  read  each  period  sej)aratelj/,  and 
give  the  name  to  each  period,  except  the  last,  or  period  of  units. 

Note. — Omit  and  in  reading  the  orders  of  units  and  periods  of  a  number. 


EXAMPLES   FOR   PRACTICE. 
Write  and  read  the  following  numbers :  — 
1    One  unit  of  the  3d  order,  two  of  the  2d,  five  of  the  1st. 
Ans.  125;  read,  one  hundred  twenty  five. 

2.  Two  units  of  the  5th  order,  four  of  the  4th,  five  of  the  2d, 
six  of  the  1st.     Ans.  24056 ;  read,  twenty-four  thousand  fiffy-six. 

3.  Seven  units  of  the  4th  order,  five  of  the  3d,  three  of  the  2d, 
eight  of  the  1st. 


NOTATION  AND  NUMERATION.  21 

4.  Nine  units  of  the  4th  order,  two  of  the  3d,  four  of  the  1st. 
6.  Five  units  of  the  4th  order,  eight  of  the  2d. 

6.  Five  units  of  the  5th  order,  one  of  the  3d,  eight  of  the  1st. 

7.  Three  units  of  the  5th  order,  six  of  the  4th,  four  of  the  3d, 
seven  of  the  1st. 

8.  Two  units  of  the  6th  order,  four  of  the  5th,  nine  of  the  4th, 
three  of  the  3d,  five  of  the  1st. 

9.  Three  units  of  the  8th  order,  five  of  the  7th,  four  of  the  6th, 
three  of  the  5th,  eight  of  the  4th,  five  of  the  3d,  eight  of  the  2d, 
seven  of  the  1st. 

10.  Three  units  of  the  9th  order,  eight  of  the  7th,  four  of  the 
6th,  six  of  the  5th,  nine  of  the  1st. 

11.  Five  units  of  the  12th  order,  three  of  the  11th,  six  of  the 
10th. 

12.  Four  units  of  the  12th  order,  five  of  the  10th,  eight  of  the 
5th,  nine  of  the  4th,  four  of  the  3d. 

13.  Three  units  of  the  15th  order,  six  of  the  14th,  five  of  the 
13th,  three  of  the  9th,  six  of  the  8th,  five  of  the  7th,  three  of  the 
3d,  six  of  the  2d,  five  of  the  1st. 

14.  Five  units  of  the  18th  order,  three  of  the  17th,  six  of  the 
16th,  four  of  the  15th,  seven  of  the  14th,  eight  of  the  13th,  four 
of  the  12th,  five  of  the  11th,  six  of  the  10th,  seven  of  the  9th, 
eight  of  the  8th,  nine  of  the  7th,  five  ot  the  6th,  six  of  the  5th, 
three  of  the  4th,  two  of  the  3d,  four  of  the  2d,  eight  of  the  1st. 

15.  Two  units  of  the  20th  order,  seven  of  the  19th,  four  of  the 
18th,  eight  of  the  13th,  five  of  the  6th,  five  of  the  5th,  five  of  the 
4th,  nine  of  the  1st. 

Write  the  following  numbers  in  figures: 

16.  Forty-eight. 

17.  One  hundred  sixty-four. 

18.  Forty-eight  thousand  seven  hundred  eighty-nine. 

19.  Five  hundred  thirty-six  million  three  hundred  forty-seven 
thousand  nine  hundred  seventy-two. 

20.  Ninety-nine  billion  thirty-seven  thousand  four. 

21.  Eight  hundred  sixty-four  billion  five  hundred  thirty-eight 
million  two  hundred  seventeen  thousand  nine  hundred  fifty-three. 


22  SIMPLE  NUMBERS. 

22.  One  hundred  seventeen  quadrillion  two  hundred  thirty-fivo 
trillion  one  hundred  four  billion  seven  hundred  fifty  million  sixty- 
six  thousand  ten. 

23.  Ninety-nine  quintillion  seven  hundred  forty-one  trillion 
fifty-four  billion  one  hundred  eleven  million  one  hundred  one. 

24.  One  hundred  octillion  one  hundred  septillion  one  hundred 
quintillion  one  hundred  quadrillion  one  hundred  trillion  one  hundred 
billion  one  hundred  million  one  hundred  thousand  one  hundred. 

25.  Four  decillion  seventy-five  nonillion  three  octillion  fifty- 
two  septillion  one  sextillion  four  hundred  seventeen  quintillion 
ten  quadrillion  twelve  trillion  fourteen  billion  three  hundred  sixty 
million  tw^enty-two  thousand  five  hundred  nineteen. 

Write  the  following  numbers  in  figures,  and  read  them : 

26.  Twenty-five  units  in  the  2d  period,  four  hundred  ninety-six 
in  the  1st.  Ans.  25,496. 

27.  Three  hundred  sixty-four  units  in  the  8d  period,  seven 
hundred  fifteen  in  the  2d,  eight  hundred  thirty-two^in  the  1st. 

28.  Four  hundred  thirty-six  units  in  the  4th  period,  twelve  in 
the  3d,  one  hundred  in  the  2d,  three  hundred  one  in  the  1st. 

29.  Eighty-one  units  in  the  5th  period,  two  hundred  nineteen 
in  the  4th,  fifty-six  in  the  2d. 

30.  Nine  hundred  forty-five  units  in  the  7th  period,  eighteen  in 
the  5th,  one  hundred  three  in  the  3d. 

31.  One  unit  in  the  10th  period,  five  hundred  thirty-six  in  the 
9th,  two  hundred  forty-seven  in  the  8th,  nine  hundred  twenty-four 
in  the  7th. 

Point  off  and  read  the  followino-  numbers : 


82. 

564. 

37. 

2005. 

33. 

24835. 

38. 

100103. 

34. 

2474783. 

39. 

53000008. 

35. 

247843112. 

40. 

1001005003. 

86. 

23678542789. 

41. 

750000040003. 

42.  247364582327896438542721. 

43.  379403270506038 

009503070. 

44.  20005700032004673000430512500000567304705030040. 


ADDITION. 


23 


ADDITION. 

69«   Addition  is  the  process  of  uniting  several  numbers  of  the 
same  kind  into  one  equivalent  number. 
-  63.   The  Sum  or  Amount  is  the  result  obtained  by  the  process 
of  addition. 

64.  When  the  given  numbers  contain  several  orders  of  units, 
the  method  of  addition  is  based  upon  the  following  principles : 

I.  If  the  like  orders  of  units  be  added  separately,  the  sum  of 
all  the  results  must  be  equal  to  the  entire  sum  of  the  given  num- 
bers.   (Ax.  10). 

II.  If  the  sum  of  the  units  of  any  order  contain  units  of  a 
higher  order,  these  higher  units  must  be  combined  with  units  of 
like  order.     Hence, 

III.  The  work  must  commence  with  the  lowest  unit,  in  ordei 
to  combine  the  partial  sums  in  a  single  expression,  at  one  ope- 
ration. 

I.  Find  the  sum  of  897,  476,  and  873. 

OPERATION  Analysis.      We   arrange   the   numbers   so   that 

897  units  of  like  order  shall  stand  in  the  same  column. 

476  We  then  add  the  first,  or  right  hand  column,  and 

o73  find  the  sum  to  be  16  units,  or  1  ten  and  6  units ; 

1746  writing  "fhe  6  units  under  the  column  of  units,  we 

add  ^e  1  ten  to  the  column  of  tens,  and  find  the 

6um  to  be  24  tens,  or  2  hundreds  and  4  tens  ;  writing  the  4  tens  under 

the  column  of  tens,  we  add  the  2  hundreds  to  the  column  of  hundreds, 

and  find  the  sum  to  be  17  hundreds,  or  1  thousand  and  7  hundreds ; 

writing  the  7  hundreds  under  the  column  of  hundreds,  and  the  1  in 

thousands*  place,  we  have  the  entire  sum,  1746. 

65.    From  these  principles  we  deduce  the  following 

HuLE.     I.   Write  the  numbers  to  he  added  so  that  all  the  units 

of  the  same  order  shall  stand  in  the  same  column)  that  is,  units 

under  units ,  tens  under  tenSy  etc, 

II.  Commencing  at  units ^  add  each  column  separately y  and  write 
the  sum  underneath^  if  it  he  less  than  ten. 


24  SIMPLE  NUMBERS. 

III.  If  the  sum  of  any  column  he  ten  or  more  than  ten,  write  the 
unit  figure  only,  and  add  the  ten  or  tens  to  the  next  column. 

IV.  Write  the  entire  sum  of  the  last  column. 

Notes. — 1.  In  adding,  learn  to  pronounce  the  partial  results  without  naming 
the  Jifjitres  separately.  Thus,  in  the  operation  given  for  illustration,  say  3,  9, 
16;  8,  15,  24,-  10,  14,  17. 

2.  When  the  sum  of  any  column  is  greater  than  9,  the  process  Of  adding  the 
tens  to  the  next  column  is  called  currying. 

OG.  Proof.  There  are  two  principal  methods  of  proving 
Addition. 

1st.  By  varying  the  combinations. 

Begin  with  the  right  hand  or  unit  column,  and  add  the  figures 
in  each  column  in  an  opposite  direction  from  that  in  which  they 
were  first  added ;  if  the  two  results  agree,  the  work  is  supposed 
to  be  right. 

2d.  By  excess  of  9's. 

07.  This  method  depends  upon  the  following  properties  of  the 
number  9 :  * 

I.  If  a  number  be  divided  by  9,  the  remainder  will  be  the  same 
as  when  the  sum  of  its  digits  is  divided  by  9.     Therefore, 

II.  If  several  numbers  be  added,  the  excess  of  9's  in  the  sum 
must  be  equal  to  the  excess  of  9's  in  the  sum  of  all  the  digits  in 
the  numbers. 

1.  Add  34852,  24784,  and  72456,  and  prove  the  work  by  the 
excess  of  9's. 

OPERATION. 

34852 

24784 

72456  ...  8,  excess  of  9's  in  all  the  digits  of  the  numbers. 
132092  ...  8,      "  "         "       "    sum        «^  " 

Analysis.  Commencing  with  the  first  number,  at  the  left  hand,  we 
say  3  and  4  are  7,  and  8  are  15  ;  dropping  9,  the  excess  is  6,  which 
added  to  5,  the  next  digit,  makes  11;  dropping  9,  the  excess  is  2; 
then  2  and  2  are  4,  and  2  (the  left  hand  digit  of  the  second  number) 

*  For  a  demonstration  of  these  properties,  see  186,  IX. 


ADDITION. 


25 


are  6,  and  4  are  10 ;  dropping  9,  the  excess  is  1.  Proceeding  in  like 
manner  through  all  the  digits,  the  final  excess  is  8  ;  and  as  8  is  also 
the  excess  of  9^s  in  the  sum,  the  work  of  addition  is  correct.  It  is 
evident  that  the  same  result  will  be  obtained  by  adding  the  digits  in 
columns  as  in  rows.  Hence,  to  prove  Addition  by  excess  of  9*s:  — 
CommenciDg  at  any  figure,  add  the  digits  of  the  given  numbers 
in  any  order,  dropping  9  as  often  as  the  amount  exceeds  9.  If 
the  final  excess  be  equal  to  the  excess  of  9^s  in  the  sum,  the  work 
is  right. 

Note. — This  method  of  proving  addition  by  the  excess  of  9's,  fails  in  tKtf- fol- 
lowing cases  :  Ist,  when  the  figures  of  the  answer  are  misplaced;  2d,  when  the 
value  of  one  figure  is  as  much  too  great  as  that  of  another  is  too  small. 

EXAMPLES   FOR   PRACTICE 


(1.) 

(2.) 

(3.) 

(4.) 

8635 

1234567 

67 

24603 

2194 

723456 

123 

298765 

7421 

34565 

4567 

47321 

5063 

45666 

89093 

•  68653 

2196 

333 

654321 

6376 

1245 

90 

1234567 

340 

26754     2038677 

5.  123+456+785+12+345+901  +  567=how  many? 

6.  r2345+67890+8763+347  +  1037  +  198760=how  many? 

7.  172+4005  +  37Gl  +  20472+367012  +  19762=how  many? 

8.  What  is  ^he  sum  of  thirty-seven  thousand  six,  four  hundred 
twenty-nine  thousand  nine,  and  two  millions  thirty-six  ? 

Ans.  2,466,051. 

9.  Add  eight  hundred  fifty-six  thousand  nine  hundred  thirty- 
three,  one  million  nine  hundred  seventy-six  thousand  eight  hun- 
dred fifty-nine,  two  hundred  three  millions  eight  hundred  ninety- 
five  thousand  seven  hundred  fifty-two.  Ans.  206,729,544. 

10.  What  is  the  sum  of  one  hundred  sixty-seven  thousand, 
three  hundred  sixty-seven  thoasand,  nine  hundred  six  thousand, 
two  hundred  forty-seven  thousand,  seventeen  thousand,  one  hun- 
dred six  thousand  three  hundred,  forty  thousand  forty-nine,  ten 
thousand  four  hundred  one  ?  Ans.  1,860,750. 

11.  What  number  of  square   miles   in  New  England,  there' 
3 


26  SIMPLE  NUMBERS. 

being  in  Maine  31766,  in  New  Hampshire  9280,  in  Vermont 
10212,  in  Massachusetts  7800,  in  Ehode  Island  1306,  and  in 
Connecticut  4674?  Ans.  65,038. 

12.  The  estimated  population  of  the  above  States,  in  1855,  was 
as  follows;  Maine  653000,  New  Hampshire  338000,  Vermont 
327000,  Massachusetts  1133123,  Ehode  Island  166500,  and  Con^ 
necticut  384000.     What  was  the  entire  population  ? 

13  At  the  commencement  of  the  year  1858  there  were  in  ope- 
ration in  the  New  England  States,  3751  miles  of  railroad;  in 
New  York,  2590  miles;  in  Pennsylvania,  2546;  in  Ohio,  2946; 
in  Virginia,  1233 ;  in  Illinois,  2678  ;  and  in  Georgia,  1233.  V>^hat 
was  the  aggregate  number  of  miles  in  operation  in  all  these  States? 

14.  The  Grand  Trunk  Railway  is  962  miles  long,  and  cost 
$60000000 ;  the  Great  Western  Eailway  is  229  miles  long,  and 
cost  S14000000;  the  Ontario,  Simcoe  and  Huron,  is  95  miles 
long,  and  cost  §3300000 ;  the  Toronto  and  Hamilton  is  38  miles 
long,  and  cost  §2000000.  What  is  the  aggregate  length,  and 
what  the  cost,  of  these  four  roads  ? 

Ans.  Length,  1,324  miles;  cost,  $79,300,000. 

15.  A  man  bequeathed  his  estate  as  follows;  to  each  of  his 
two  sons,  $12450;  to  each  of  his  three  daughters,  $6500;  to 
his  wife,  $650  more  than  to  both  the  sons,  and  the  remainder, 
which  was  $1000  more  than  he  had  left  to  all  his  family,  he  gave 
to  benevolent  institutions.  What  was  the  whole  amount  of  hig 
property?  A71S,  $140,900. 

16.  How  many  miles  from  the  southern  extremity  of  Lake 
Michigan  to  the  Gulf  of  St.  Lawrence,  passing  through  Lake 
Michigan,  330  miles ;  Lake  Huron,  260  miles ;  Eiver  St.  Clair,  24 
miles;  Lake  St.  Clair,  20  miles;  Detroit  River,  23  miles;  Lake 
Erie,  260  miles;  Niagara  River,  34  miles;  Lake  Ontario,  180 
miles;  and  the  River  St.  Lawrence,  750  miles? 

17.  The  United  States  exported  molasses,  in  the  year  1856,  to 
the  value  of  $154630;  in  1857,  $108003;  in  1858,  $115893; 
and  tobacco,  during  the  same  years  respectively,  to  the  value  of 
$1829207,  $1458553,  and  $2410224.  What  was  the  entire  valua 
of  the  molasses  and  tobacco  exported  in  these  three  years  ? 


ADDITION. 


27 


18.  The  population  of  Boston,  in  1855,  was  162629;  Provi- 
dence, 50000;  New  York,  629810;  Philadelphia,  408815;  Brook- 
lyn, 127618  ;  Cleveland,  43740 ;  and  New  Haven,  25000.  What 
•was  the  entire  population  of  these  cities  ?  Ans.  1,447,612. 

19.  Iron  was  discovered  in  Greece  by  the  burning  of  Mount 
Ida,  B.  C.  1406;  and  the  electro-magnetic  telegraph  w^as  invented 
by  Morse,  A.  D.  1832.  What  period  of  time  elapsed  between  the 
two  events  ?  Ans.  3,238  years. 

20.  The  number  of  pieces  of  silver  coin  made  at  the  United 
States  Mint  at  Philadelphia,  in  the  year  1858,  were  as  follows : 
4628000  half  dollars,  10600000  quarter  dollars,  690000  dimes, 
4000000  half  dimes,  and  1266000  three-cent  pieces.  What  was 
the  total  number  of  pieces  coined  ? 

21.  The  cigars  imported  by  the  United  States,  in  the  year  1856, 
were  valued  at  63741460;  in  1857,  at  $4221096;  and  in  1858,  at 
$4123208.  What  was  the  total  value  of  the  importations  for  the 
three  years  ?  Ans.  $12,085,764. 

22.  In  the  appropriations  made  by  Congress  for  the  year  ending 
June  30,  1860,  were  the  following;  for  salary  and  mileage  of 
members  of  Congress,  $1557861;  to  officers  and  clerks  of  both 
Houses,  $157639 ;  for  paper  and  printing  of  both  Houses,  $170000 ; 
to  the  President  of  the  United  States,  $31450 ;  and  to  the  Yice 
President,  $8000.     What  is  the  total  of  these  items  ? 


ADDING   TWO    OR    MORE   COLUMNS  AT    ONE    OPERATION. 

68,   1.  What  is  the  sum  of  4632,  2553,  4735,  and  2863  ? 

OPERATION.  Analysis.     Beginning  with  the  units  and  tens  of 

4632  the  number  last  written,  we  add  first  the  tens  above, 

2553  then  the  units,  thus  ;  63  and  30  are  93,  and  5  are  98, 

4735  and  50  are  148,  and  3  are  151,  and  30  are  181,  and 

__.__  ^  ^^^  ^^^'     ^^  *^^^  sum,  we  write  the  83  under  the 

14783  columns  added,  and  carry  the  1  to  the  next  columns, 

thus ;  28  and  1  are  29,  and  40  are  G9,  and  7  are  76, 

and  20  are  96,  and  5  are  101,  and  40  are  141,  and  6  are  147,  which 

we  write  in  its  place,  and  we  have  the  whole  amount,  14783. 


28 


SIMPLE  NUMBERS. 

EXAMPLES 

FOR  PRACTICE. 

-0 

(1.) 

(2.) 

(3.) 

(4.) 

8450 

75634 

123456 

7349042 

5425 

86213 

47021 

2821986 

8595 

92045 

82176 

1621873 

6731 

73461 

570914 

236719 

7963 

34719 

379623 

401963 

5143 

26054 

7542 

67254 

4561 

19732 

25320 

45067 

6783 

84160 

57644 

910732 

4746 

97013 

908176 

6328419 

2373 

34567 

73409 

1437651 

8021 

43651 

3147 

9716420 

7273 

52170 

67039 

8191232 

71064 

719419 

2345467 

34128358 

5.  What  is  the  total  number  of  churches,  the  number  of  persons 
accommodated,  and  the  value  of  church  property  in  the  United 
States,  as  shown  by  the  following  statistics  ? 

No.  of  No.  of  persons  Value  of 

churches.  accommodated.  church  property 

Methodist 12484  4220293  $14636671 

Baptist 8798  3134438  10931382 

Presbyterian 4591  2045516  14469889 

Congregational 1675  795677  7973962 

Episcopal 1430  631613  11261970 

Koman  Catholic 1269  705983  8973838 

Lutheran 1205  532100  2867886 

Christians 812  296050  845810 

Friends 715  283023  1709867 

Union 619  213552  690065 

Universalist 494  205462  1767015 

Free  Church 361  108605  252255 

Moravian :...      331  112185  443347 

German  Reformed 327  156932  965880 

Dutch  Reformed 324  181986  4096730 

Unitarian 244  137867  8268122 

Mennonite 110  29900  94245 

Tunkers 52  35075  46025 

Jewish... 31  16575  371600 

Swedenborgian 15  5070  108100 


ADDITION.  29 


I^H    6.  Give  the  amounts  of  the  productions  of  the  United  States  and 
Territories  for  the  year  1850,  as  expressed  in  the  following  columns : 

Pounds  of 
Butter. 

Alabama 4,008,811 

Arkansas 1,854,239 

California 705 

Columb.,Dist.  14,872 
Connecticut...    6,498,119 

Delaware 1,055,308 

Florida 371,498 

Georcria 4,640,559 

Illinois 12,526,543 

Indiana 12,881,535 

Iowa 2,171,188 

Kentucky....  9,947,523 
Louisiana...'..       683,069 

Maine 9,243,811 

Maryland 3,086,160 

Massachusetts  8,071,370 

Michigan 7,065,878 

Mississippi...    4,346,234 

Missouri 7,834,359 

N.Hampshire  6,977,056 
New  Jersey...  9,487,210 
New  York....  79,766,094 
N.Carolina...    4,146,290 

Ohio 34,449,379 

Pennsylvania  39,878,418 
llhode  Island  995,670 
S.  Carolina...  2,981,850 
Tennessee....    8,139,585 

Texas 2,344,900 

Vermont 12,137,980 

Virginia 11,089,359 

Wisconsin....  3,633,750 
Territories...       295,984 


gi* 


Pounds  of 
Cheese. 

Pounds  of 
Wool. 

Bushels  of 
Wheat. 

31,412 

657,118 

294,044 

30,088 

182,595 

199,639 

150 

5,520 

17,228 

1,500 

525 

17,370 

5,363,277 

497,454 

41,762 

3,187 

57,768 

482,511 

18,015 

23,247 

1,027 

46,976 

990,019 

1,088,534 

1,278,225 

2,150,113 

9,414,575 

624,564 

2,610,287 

6,214,458 

209,840 

373,898 

1,530,581 

213,954 

2,297,433 

2,142,822 

1,957 

109,897 

417 

2,434,454 

1,364,034 

296,259 

3,975 

477,438 

4,494,680 

7,088,142 

585,138 

31,211 

1,011,492 

2,043,283 

4,925,889 

21,191 

559,619 

137,990 

203,572 

1,627,164 

2,981,652 

3,196,563 

1,108,476 

185,658 

365,756 

375,396 

1,601,190 

49,741,413 

10,071,301 

13,121,498 

95,921 

970,738 

2,130,102 

20,819,542 

10,196,371 

14,487,351 

2,505,034 

4,481,570 

15,367,691 

316,508 

129,692 

49 

4,970 

487,233 

1,066,277 

177,681 

1,364,378 

1,619,386 

95,299 

131,917 

41,729 

8,720,834 

3,400,717 

535,955 

436,292 

2,860,765 

11,212,616 

400,283 

253,963 

4,286,131 

73,826 

71,894 

517,562 

80  SIMPLE  NUMBERS. 


SUBTRACTION. 

60.  Subtraction  is  the  process  of  determining  the  difference, 
between  two  numbers  of  the  same  unit  value. 

•J'O,    The  Minuend  is  the  number  to  be  subtracted  from. 

yi.    The  Subtrahend  is  the  number  to  be  subtracted. 

7^.  The  Difference  or  Eemainder  is  the  result  obtained  by 
the  process  of  subtraction. 

73.  When  the  given  numbers  contain  more  than  one  figure 
each,  the  method  of  subtraction  depends  upon  the  following  prin- 
ciples : 

I.  If  the  units  of  each  order  in  the  subtrahend  be  taken  sepa- 
rately  from  the  units  of  like  order  in  the  minuend,  the  sum  of  the 
differences  must  be  equal  to  the  entire  difference  of  the  given 
numbers.     (Ax.  10 .) 

II.  If  both  minuend  and  subtrahend  be  equally  increased^  the 
remainder  will  not  be  changed. 

1.    From  928  take  275. 

OPERATION.  Analysis.     We  first  subtract  5  units  from 

Minuend,  928  8  units,  and  obtain  3  units  for  a  partial  re- 

Subtraheud,       275  mainder.     As  we  cannot  take  7  tens  from  2 

Remainder  653  tens,  we  add  10  tens  to  the  2  tens,  making 

12  tens ;  then  7  tens  from  12  tens  leave  5 
tens,  the  second  partial  remainder.  Now,  since  we  added  10  tens, 
or  1  hundred,  to  the  minuend,  we  will  add  1  hundred  to  the  subtra- 
hend, and  the  true  remainder  will  not  be  changed  (II) ;  thus,  1 
hundred  added  to  2  hundreds  makes  3  hundreds,  and  this  sum  sub- 
tracted from  9  hundreds  leaves  6  hundreds ;  and  we  have  for  the  total 
remainder,  653. 

NoTTi;. — The  process  of  adding  10  to  the  minuend  is  sometimes  called  hor- 
rowwrj  10,  and  that  of  adding  1  to  the  next  figure  of  the  subtrahend,  carrying  1. 

I' J:.  From  these  principles  and  illustrations  we  deduce  the 
following 

HuLE.  I.  Write  the  less  number  under  the  greater.^  'placing  units 
of  the  same  order  under  each  other. 


SUBTRACTION.  gj 

II.  Begin  at  the  right  Tiandj  and  taJce  each  figure  of  the  suhtra- 
hend  from  the  figure  above  it,  and  ■write  the  remit  underneath. 

III.  If  any  figure  in  the  subtrahend,  he  greater  than  the  corres- 
pond ing  figure  above  it^  add  10  to  that  upper  figure  before  sub- 
tractingj  and  then  add  1  to  the  next  left  hand  figure  of  the  subtra- 
(lend. 

7^*  Proof.  It  is  evident  that  the  subtrahend  and  remainder 
must  together  contain  as  many  units  as  the  minuend ;  hence,  to 
prove  subtraction,  we  have  three  methods  : 

1st.  Add  the  remainder  to  the  subtrahend;  the  sum  will  be 
equal  to  the  minuend.      Or, 

2d.  Subtract  the  remainder  from  the  minuend ;  the  difference 
will  be  equal  to  the  subtrahend.     Or, 

3d.  Find  the  excess  of  9's  in  the  remainder  and  subtrahend 
together,  and  it  will  be  equal  to  the  excess  of  9's  in  the  minuend. 

EXAMPLES   FOR   PRACTICE. 

(1.)      (2.)       (3.)       (4.) 
From    47965    103767    57610218   '89764321 
Take    26714     98731     8306429    83720595 

Eem.    21251     5036    49303789    6043726 

5.  From  180037561  take  5703746. 

6.  From  2460371219  take  98720342. 

7.  89037426175  —  2435036749  =  how  many? 

8.  10000033421  —  999044110  =  how  many? 

9.  A  certain  city  contains  146758  inhabitants,  which  is  3976 
more  than  it  contained  last  year;  how  many  did  it  contain  last 
year?  Ans.  142,782. 

10.  The  first  newspaper  published  in  America  was  issued  at 
Boston  in  1704;  how  long  was  that  before  the  death  of  Benjamin 
Franklin,  which  occurred  in  1790  ? 

11.  A  merchant  sold  a  quantity  of  goods  for  $42017,  which 
was  $1675  more  than  they  cost  him;  how  much  aid  they  cost 
him?  Ans.  $40,342. 

12.  In  1858  the  exports  of  the  United  States  amounted  to 


82  SIMPLE  NUMBERS. 

$324644421,  and  the  imports  to  $282613150;  how  much  did  the 
exports  exceed  the  imports  ?  Ans.  $42,031,271. 

13.  In  1858  the  gold  coinage  of  the  United  States  amounted  to 
$52889800,  and  the  silver  to  $8233287;  how  much  did  the  gold 
exceed  the  silver  coinage  ? 

14.  The  South  in  1850  produced  978311690  pounds  of  cotton, 
valued  at  $101834616,  and  237133000  pounds  of  sugar  valued  at 
$16599310;  how  much  did  the  cotton  exceed  the  sugar  in  quan- 
tity and  in  value  ?         Ans.  741,178,690  pounds;  $85,235,306. 

15.  The  area  of  the  Chinese  Empire  is  5110000  square  miles, 
and  that  of  the  United  States  2988892  square  miles ;  the  esti- 
mated population  of  the  former  is  340000000,  and  that  of  the 
latter  in  1850  was  23363714.  What  is  the  difference  in  area  and 
in  population  ? 

16.  The  population  of  London  in  1850  was  2362000,  and  that 
of  New  York  city  515547;  how  many  more  inhabitants  had  London 
than  New  York  ?  A7is.     1,846,453. 

17.  The  total  length  of  railroads  in  operation  in  the  United 
States,  January  1,  1859,  was  27857  miles,  and  the  total  length  of 
the  canals  5131  miles;  how  many  miles  more  of  railroad  than  of 
canal?  Ans.  22,726. 

18.  The  entire  deposit  of  domestic  gold  at  the  United  States 
Mint  and  its  branches,  to  June,  1859,  was  $470341478,  of  which 
$451310840  was  from  California;  how  much  was  received  from 
other  sources  ?  Ans.  $19,030,638. 

19.  During  the  year  ending  September  30,  1858,  the  number 
of  letters  exchanged  between  the  United  States  and  Great  Britain 
were  1765015  received,  and  1603609  sent;  between  the  United 
States  and  France,  624795  received,  and  639906  sent.  How  many 
letters  did  the  exchange  with  Great  Britain  exceed  those  with 
France?  Ans.  2,103,923. 

20.  The  Southern  States  in  1850  had  a  population  of  6696061, 
the  Middle  States  6624988,  and  the  Eastern  States  2728116; 
how  many  more  inhabitants  had  the  Middle  and  Eastern  States 
than  the  Southern  States  ? 

21.  Having  $20000,  I  wish  to  know  how  much  more  I  must 


SUBTRACTION. 


33 


accumulate  to  be  able  to  purchase  a  piece  of  property  worth  $23470, 
and  have  $5400  left?  Ans.  ^8,870. 

22.  A  has  §3540  more  than  B,  and  §1200  less  than  C,  who  has 
S20600 ;  D  has  as  much  as  A  and  B  together.     How  much  has  D  ? 

Ans.  §35,260. 

TWO    OR    MORE    SUBTRAHENDS. 

•yO.  Two  or  more  numbers  may  be  taken  from  another  at  a 
single  operation,  as  shown  by  the  following  example  : 

I.  A  man  having  1278  barrels  of  flour,  sold  236  barrels  to  A, 
362  to  B,  and  387  to  C;  how  many  had  he  left? 

OPERATION.  Analysis.     Since  the  remainder  sought, 

1278  added  to  the  subtrahends,  must  be  equal  to 

the  minuend,  we  add  the  columns  of  the 
subtrahends,  and  supply  such  figures  in  the 
remainder  as,  combined  with  these  sums, 
will  produce  the  minuend.  Thus,  7  and  2 
are  9,  and  6  are  15,  and  3  (supplied  in  the 
remainder  sought)  are  18 ;  then,  carrying 
the  tens^  figure  of  the  18,  1  and  8  are  9,  and  6  are  15,  and  3  are  18, 
and  9  (supplied  in  the  remainder)  are  27;  lastly,  2  to  carry  to  3  are 
5,  and  3  are  8,  and  2  are  10,  and  2  (supplied  in  the  remainder)  are 
12  ;  and  the  whole  remainder  is  293.     Hence,  the  following 

BuLE.  T.  Hamng  written  the  several  subtrahends  under  tJi^-  min- 
uend j  add  the  first  coIu7nn  of  the  subtrahends,  and  supply  such  a 
figure  in  the  remainder  sought^  as,  added,  to  this  partial  sum,  will 
give  an  amount  having  for  its  unit  figure  the  figure  above  in  the 
minuend. 

II,  Carry  the  tens^  figure  of  this  amount  to  the  next  column  of 
the  subtrahends  J  and  proceed  as  before  till  the  entire  re.main<der  is 
obtained. 

EXAMPLES    FOR   PRACTICE. 


Minuend, 


Subtrahends, 


Remainder, 


From 

47962 

(2.) 
127368 

(3.) 
903486 

(4.) 
2503734 

Take  j 

21435 

15672 

456 

56304 

4782 
9156 

430164 

182875 

67321 

89763 

94207 

237564 

Rem. 

10399 

57126 

273126 

2082200 

34  SIMPLE  NUMBERS. 

5.  From  4568  take  1820  +  275  +  320. 

6.  Subtract  1200  +  750  -j-  96  from  4756  +  575  +  140  +  84. 

7.  A  man  bought  four  city  lots,  for  which  he  paid  $15760.  For 
the  first  he  paid  §2175,  for  the  second  $3794,  and  for  the  third 
$4587;  how  much  did  he  pay  for  the  fourth  ?       Aiis.  $5,204. 

8.  John  Wise  owns  property  to  the  amount  of  $75860^  of  which 
he  has  §45640  invested  in  real  estate,  $25175  in  personal  property, 
and  the  remainder  he  has  in  bank ;  how  much  has  he  in  bank  ? 

9.  Lake  Huron  contains  20000  square  miles;  by  how  much 
does  it  exceed  the  area  of  Lake  Erie  and  Lake  Ontario,  the  former 
containing  11000  square  miles,  and  the  latter  7000  ? 

Ans.  2000  square  miles. 

10.  In  the  year  1852,  there  arrived  in  the  United  States  398470 
immigrants,  of  whom  157548  were  born  in  Ireland,  and  143429 
were  born  in  Germany;  how  many  were  born  in  other  countries? 

Ajis.  97,493. 

11.  The  entire  amount  of  coinage  in  the  United  States  for  the 
year  ending  June,  1858,  was  $61357088,  of  which  $52889800  was 
of  gold,  $234000  of  copper,  and  the  remainder  of  silver;  how  much 
was  of  silver  ? 

12  A  speculator  gained  $5760,  and  afterward  lost  $2746;  at 
another  time  he  gained  $3575,  and  then  lost  $4632.  How  much 
did  his  gains  exceed  his  losses?  Ans.  $1,957. 

13.  The  Eastern  States  have  an  area  of  65038  square  miles, 
the  Middle  States  114624  square  miles,  and  the  Southern  States 
643166  square  miles;  how  many  more  square  miles  have  the 
Southern  than  the  Middle  and  Eastern  States  ? 

14.  The  entire  revenue  of  the  United  States  Post  Office  Depart- 
ment for  the  year  ending  Sept.  30,  1858,  was  $8186793,  of  which 
sum  $5700314  was  received  for  stamps  and  stamped  letters,  and 
$904299  for  letter-postage  in  money;  how  much  was  received  from 
all  other  sources  ?  ^ns.  $1,582,180. 

15.  The  total  expenditures  ot  the  Department  for  the  same  year 
were  §12722470,  of  which  sum  $7821556  was  paid  for  the  trans- 
portation of  inland  mails,  $424497  for  the  transportation  of  foreign 
mails  and  $2355016  as  compensation  to  postmasters;  how  much 
was  expended  for  all  other  purposes?  $2,121^401. 


MULTIPLICATION. 


M 


MULTIPLICATION. 

77a  Multiplication  is  the  process  of  taking  one  of  two  given 
numbers  as  many  times  as  there  are  units  in  the  other, 

7S»    The  Multiplicand  is  the  number  to  be  taken. 

7@,  The  Multiplier  is  the  number  which  shows  how  many 
times  the  multipHcand  is  to  be  taken. 

§®.-  The  Product  is  the  result  obtained  by  the  process  of  mul- 
tiplication. 

8S.   The  Factors  are  the  multiplicand  and  multiplier. 

Is'oT[-:s. —  1.  Factors  are  producers,  and  the  multiplicand  and  multiplier  are 
called  iMCtors  because  tbey  produce  the  product. 

2.  MuUiplication  is  a  short  method  of  performing  addition  when  the  numbers 
to  be  jidded  are  equal. 

§S,  The  method  of  multiplying  when  either  factor  contains 
more  than  one  figure,  depends  upon  the  following  principles  : 

It  is  evident  that  5  units  taken  3  times  is  the  same  as  3  units 
taken  5  times ;  and  the  same  is  true  of  any  two  factors.     Hence, 

I.  The  product  of  any  two  factors  is  the  same,  whichever  is  used 
as  the  multiplier.  If  units  be  multiplied  by  units,  the  product 
will  be  units ;  if  tens  be  multiplied  by  units,  or  units  by  tens,  the 
product  will  be  tens;  and  so  on.     That  is, 

II.  If  either  factor  be  units  of  the  first  order,  the  product  will 
be  units  of  the  same  order  as  the  other  flictor. 

III.  If  the  units  of  each  order  in  the  multiplicand  be  taken  sepa- 
rately as  many  times  as  there  are  units  in  the  multiplier,  the  sum 
of  the  products  must  be  equal  to  the  entire  product  ©f  the  given 
numbers,     (Ax.  10). 

1.   Multiply  346  by  8. 

OPERATION.  Analysis.     In   this    example   it  is   re- 

Multiplicand,  346  ej[uired  to  take  346  eight  times.     If  we  take 

Multiplier,  8  the  units  of  each  order  8  times,  we  shall 

Pj.q^^^^  276R  *^^^®   ^^^®   entire   number   8   times,  (III). 

Therefore,  commencing  at  the  right  hand. 

we  say;  8  times  6  units  are  48  units,  or  4  tens  and  8  units  ;  writing 

the  8  units  in  the  product  in  units'  place,  we  reserve  the  4  tens  to 

add  to  the  next  product ;  8  times  4  tens  are  32  tens,  and  the  4  tens 


J0  SIMPLE  NUMBERS. 

reserved  in  the  last  product  added,  are  3G  tens,  or  3  hundreds  and  6 
tens ;  we  write  the  6  tens  in  the  product  in  tens^  place,  and  reserve 
the  3  hundreds  to  add  to  the  next  product ;  8  times  3  hundreds  are 
24  hundreds,  and  the  3  hundreds  reserved  in  the  last  product  added, 
are  27  hundreds,  which  being  written  in  the  product,  each  figure  in 
the  place  of  its  order,  gives,  for  the  entire  product,  2768. 

2.  Multiply  758  by  346. 

OPERATION.  Analysis.     In  this  example  the  multiplicand  is 

758  t^  be  taken  346  times,  which  may  be  done  by 

346  taking  the  multiplicand  separately  as  many  tinues 

TZTo  as  there  are  units  expressed  by  each  figure  of  the 

8032  multiplier.     758   multiplied   by  6   units   is  4548 

2274  units,  (II) ;  758  multiplied  hj  4  tens  is  3032  tens, 

209900.  (^I)'  which  we  write  with  its  lowest  order  in  tens' 

place,  or  under  the  figure  used  as  a  multiplier; 

758  multiplied  by  3  hundreds  is  2274  hundreds,  (II),  which  we  write 

with  its  low3st  order  in  hundreds^  place.     Since  the  sum  of  these 

products  must  be  the  entire  product  of  the  given  numbers,  (III),  we 

add  the  results,  and  obtain  262268,  the  answer. 

Notes. — 1.  When  the  multiplier  contains  two  or  more  figures,  the  several  re- 
sults obtained  by  multiplying  lay  each  figure  are  called  partial  pmdxictn. 

2.  When  there  are  ciphers  between  the  significant  figures  of  the  multiplier, 
j)ass  over  them,  and  multiply  by  the  significant  figures  only. 

83.  From  these  principles  and  illustrations  we  deduce  the  fol- 
lowing general 

Rule.  I.  Write  the  multiplier  under  the  midtifpUcandy  jplacmg 
units  of  the  same  order  under  each  other. 

II.  Multiply  the  multiplicand  hy  each  figure  of  the  multiplier 
successively  J  beginning  with  the  unit  figure,  and  write  the  first  figure 
of  each  partial  product  under  the  figure  of  the  multiplier  used, 
writing  doivn  and  carrying  as  in  addition. 

III.  If  there  are  partial  products,  add,  them,  and  their  sum  will 
he  the  product  required. 

Note.  —The  multiplier  denotes  simply  the  number  of  times  the  multiplicand 
is  to  be  taken  -,  hence,  in  the  analysis  of  a  problem,  the  multiplier  must  be  con. 
sidered  as  abstract,  though  the  multiplicand  may  be  either  abstract  or  concrete. 

84,  Proof.  There  are  two  principal  methods  of  proving 
multiplication 


MULTIPLICATION.  37 

1st.  By  varying  the  partial  products. 

Invert  the  order  of  the  factors ;  that  is,  multiply  the  multiplier 
by  the  multiplicand :  if  the  product  is  the  same  as  the  first  result, 
the  work  is  correct. 

2d.  By  excess  of  9's. 

85.  The  illustration  of  this  method  depends  upon  the  following 
principles  : 

I.  If  the  excess  of  9's  be  subtracted  from  a  number^  the  re- 
mainder will  be  a  number  having  no  excess  of  9's. 

II.  If  a  number  having  no  excess  of  9's  be  multiplied  by  any 
number,  the  product  will  have  no  excess  of  9's. 

1.  Let  it  be  required  to  multiply  473  by  138. 

OPERATION.  Analysis.     The    excess   of 

473  =  468  +  5  9's  in  473  is  5,  and  473  =  468 

138  =  135  +  3  +  5,  of  which  the  first  part, 

)468  v~T35  =  63180        ^^^^  contains  no  excess  of  9's, 
5x135=      675         W'      The   excess   of   O's    in 
468  X  3      =    1404        138  is  3,  and  138  =  135  +  3,  of 
5x3      =         15         which  the  first  part,  135,  con- 
Entire  product,  65274        tains  no   excess    of   9's,    (I). 

Multiplying  both  parts  of  the 
multiplicand  by  each  part  of  the  multiplier,  we  have  four  partial  pro- 
ducts, of  which  the  first  three  have  no  excess  of  9's,  because  each  con- 
tains a  factor  having  no  excess  of  9's,  (II).  Therefore,  the  excess 
of  9's  in  the  entire  product  must  be  the  same  as  the  excess  of  9's  in 
the  last  partial  product,  15,  which  we  find  to  be  1  -f-  5  =  6.  The  same 
may  be  shown  of  any  two  numbers.  Hence,  to  prove  multiplication 
by  excess  of  9's, 

Find  the  excess  of  9's  in  each  of  the  two  factors,  and  multiply 
them  together;  if  the  excess  of  9's  in  this  product  is  equal  to  the 
excess  of  9's  in  the  product  of  the  factors,  the  work  is  supposed 
to  be  right. 

NoTK. — If  the  excess  of  9's  in  either  factor  is  0,  the  excess  of  9*s  in  the  pro- 
duct will  be  0,  (II). 

EXAMPLES    FOR    PRACTICE. 

(1.)  (2.)                  (3.)  (4.) 

Multiply     475  3172                9827  7198 

By                   9  14  84  216 

Prod.        4275  4S08  82546B  1554768 
4 


88  SIMPLE  NUMBERS. 

5.  Multiply  81416  by  175.  Ans,  5497800. 

6.  Multiply  40930  by  779.  Ajis.  31884470. 

7.  Multiply  46481  by  936. 

8.  Multiply  15607  by  3094. 

9.  Multiply  281216  by  978.  Ans.  275029248. 

10.  Multiply  30204  by  4267.  Ans.  128,880,468. 

11.  What  is  the  product  of  4444  x  2341  ? 

A71S.  10,403,404. 

12.  What  is  the  product  of  4567  X  9009  ? 

Ans.  41,144,103: 

13.  What  is  the  product  of  2778588  x  9867? 

Ans.  27,416,327,796. 

14.  What  is  the  product  of  7060504  x  30204  ? 

Ans.  213,255,462,816. 

15.  AYhat  will  be  the  cost  of  building  276  miles  of  railroad  at 
$61320  per  mile  ?  Ans.   $16,924,320. 

16.  If  it  require  125  tons  of  iron  rail  for  one  mile  of  railroad, 
how  many  tons  will  be  required  for  196  miles  ? 

17.  A  merchant  tailor  bought  36  pieces  of  broadcloth,  each 
piece  containing  47  yards,  at  7  dollars  a  yard ;  how  much  did  he 
pay  for  the  whole  ?  Ans.   $11,844. 

18.  The  railroads  in  the  State  of  New  York,  in  operation  in 
1858,  amounted  to  2590  miles  in  length,  and  their  average  cost 
was  about  $52916  per  mile ;  what  was  the  total  cost  of  the  rail- 
roads in  New  York  V  Ans.  $187,052,440. 

19.  The  Illinois  Central  Eailroad  is  700  miles  long,  and  cost 
§45210  per  mile;  w^hat  was  its  total  cost? 

20.  The  salary  of  a  member  of  Congress  is  $3000,  and  in  1860 
there  were  303  members;  how  much  did  they  all  receive  ? 

21.  The  United  States  contain  an  area  of  2988892  square  miles, 
and  in  1850  they  contained  8  inhabitants  to  each  square  mile; 
what  was  their  entire  population?  Ans.   23,911,186. 

22.  Great  Britain  and  Ireland  have  an  area  of  118949  square 
miles,  and  in  1850  they  contained  a  population  of  232  to  the  square 
mile;  what  was  their  entire  population?  Ans.   27,596,168. 

23.  The  national  debt  of  France  amounts  to  $32  for  each  indi- 


MULTIPLICATIOJT. 


39 


Tidual,  and  the  population  in  1850  was  35781628 ;  what  was  the 
entire  debt  of  France  ?  Ans.  1,145,012,096. 

POWERS    OF    NUMBERS. 

§6.  We  have  learned  (1^)  that  a  power  is  the  product  arising 
from  multiplying  a  number  by  itself,  or  repeating  it  any  number 
of  times  as  a  factor;  (1^),  that  a  root  is  a  factor  repeated  to  pro- 
duce a  power;  and  (40)  an  index  or  exponent  is  the  number  in- 
dicating the  power  to  which  a  number  is  to  be  raised. 

S7,  The  First  Power  of  any  number  is  the  number  itself,  or 
the  root;  thus,  2,  3,  5,  are  first  powers  or  roots. 

§§.  The  Second  Power,  or  Square,  of  a  number  is  the  pro- 
duct arising  from  using  the  number  two  times  as  a  factor;  thus, 
22^2x2  =  4;  52=5x5=:.25. 

§9.  The  Third  Power,  or  Cube,  of  a  number  is  the  product 
arising  from  using  the  number  three  times  as  a  factor;  thus, 
43=4  X  4  X  4  =  64. 

9©.  The  higher  powers  are  named  in  the  order  of  their  num- 
bers, as  Fourth  Power ^  Fifth  Power ^  Sixth  Poioer,  etc. 

91a    1.  What  is  the  third  power  or  cube  of  23  ? 

OPERATION.  Analysis.     We   multiply  23 

23  X  23  X  23  =  12167  ^J  ^3,  and  the  product  by  23; 

and,  since  23  has  been  taken  3 
times  as  a  factor,  the  last  product,  121G7,  must  be  the  third  power  or 
cube  of  23.     Hence, 

Rule.  Multiphj  the  numher  ly  itself  as  'many  times,  less  1,  as 
there  are  units  in  the  exponent  of  the  required  poicer. 

ISToTK. — The  process  of  producing  any  required  power  of  a  number  by  multi- 
plication is  called  Involution, 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  square  of  72  ?  Ans    5184. 

2.  What  is  t\iQ  fifth  power  of  12  ?  Ans.  248832. 

3.  What  is  the  cube  of  25  ? 

4.  What  is  the  seventh  power  of  7  ?  Ans,  823543. 


40  SIMPLE  NUMBERS. 

6.  What  is  the  fourth  power  of  19  ?  Ans.  130321. 

6.  Required  the  sixth  power  of  3.  Aiis.  729. 

7.  Find  the  powers  indicated  in  the  following  expr^^ssions : 
9^  IP,  18^  125S  786^  94^  100*,  17»,  251.' 

8.  Multiply  8»  by  15^  A)is.  115200. 

9.  What  is  the  product  of  25^  X  3*? 

10.  7'  X  200  =  4*  X  ll^  and  how  many  ?  Ans.  37.624. 

GENERAL  PRINCIPLES  OF  MULTIPICATION. 

02.  There  are  certain  general  principles  of  multiplication,  of 
use  in  various  contractions  and  applications  which  occur  in  sub- 
sequent portions  of  this  work.  These  relate,  1st,  to  changing  the 
factors  by  addition  or  subtraction;  2d,  to  the  use  of  successive 
factors  in  continued  multiplication. 

CHANGING    THE    FACTORS    BY    ADDITION    OR    SUBTRACTION. 

93.  The  product  is  equal  to  either  factor  taken  as  many  times 
as  there  are  units  in  the  other  factor.     (8S,  I).     Hence, 

I.  Adding  1  to  either  /actor,  adds  the  other  /actor  to  the  pro- 
duct. 

11.  SubtrcLctlng  1  /rom  either  /actor ,  subtracts  the  other  /actor 
/rom  the  pj^oduct.     Hence, 

III.  Adding  an?/  number  to  either  /actor,  increases  the  pro-, 
duct  by  as  many  times  the  other  /actor  as  there  are  units  in  the 
number  added;  and  SUBTRACTING  any  number  /rom  either  /actor, 
DIMINISHES  the  product  by  as  many  times  the  other /actor  as  there 
are  units  in  the  number  subtracted. 

CONTINUED  MULTIPLICATION. 

94.  A  Continued  Multiplication  is  the  process  of  finding  the 
product  of  three  or  more  factors,  by  multiplying  the  first  by  the 
second,  this  result  by  the  third,  and  so  on. 

95.  To  show  the  nature  of  continued  multiplication,  we 
observe : 

1st.  If  any  number,  as  17,  be  multiplied  by  any  other  number, 
as  3,  the  result  will  be  3  times  17;  if  this  result  be  multiplied  by 


MULTIPLICATION.  41 

another  number,  as  5,  the  new  product  will  be  5  times  3  times  17, 
which  is  evidently  15  times  17.  Hence,  17  X  3  x  5  =  17  X  15; 
the  same  reasoning  would  extend  to  three  or  more  multipliers. 

2d.  Since  5  times  3  is  equal  to  3  times  5,  (82, 1),  it  follows  that 
17  multiplied  by  5  times  3  is  the  same  as  17  multiplied  by  3  times 
5 ;  or  17  X  3  X  5  =  17  X  5  X  3.  Hence,  the  product  is  not 
changed  by  changing  the  orders  of  the  factors. 

These  principles  may  be  stated  as  follows : 

I.  If  a  given  number  be  multiplied  by  several  factors  in  con- 
tinued multiplication,  the  result  will  be  the  same  as  if  the  given 
number  were  multiplied  by  the  product  of  the  several  multipliers. 

II.  The  product  of  several  factors  in  continued  multiplication 
will  be  the  same,  in  whatever  order  the  factors  are  taken. 

CONTRACTIONS  IN  MULTIPLICATION. 
CASE  I. 

96.  "When  the  multiplier  is  a  composite  number. 

A  Composite  Number  is  one  that  may  be  produced  by  multi- 
plying together  two  or  more  numbers.  Thus,  18  is  a  composite 
number,  since  6  X  3  =  18 ;  or,  9  X  2  =  18 ;  or,  3  X  3  X  2=18. 

97.  The  Component  Factors  of  a  number  are  the  several 
numbers  which,  multiplied  together,  produce  the  given  number; 
thus,  the  component  factors  of  20  are  10  and  2  (10  X  2  =  20); 
or,  4  and  5  (4  X  5  =  20);  or,  2  and  2  and  5  (2  X  2  x  5  =  20). 

Note. — The  pupil  must  not  confound  the /ac/ors  with  the  parta  of  a  number. 
Thus,  the  factors  of  which  12  is  composed,  are  4  and  3  (4X3  =  12) ;  while  the 
parts  of  which  12  is  composed  are  8  and  4  (8  +  4=12);  or  10  and  2  (10  +  2  =  12). 
The/acfors  are  multiplied,  while  ih.Q  parts  are  added,  to  produce  the  number. 

98.  1.    Multiply  327  by  35. 

OPERATION. 

327  Analysis.     The  factors  of  35  are  7  and  5. 

7  We  multiply  327  by  7,  and  this  result  by  5, 

ooQQ  and  obtain  11445,  -which  must  be  the  same 

5  as  the  product  of  327  by  5  times  7,  or  35. 

(95,  I).     Hence  we  have  the  following 


11445 
4* 


42  SIMPLE  NUMBERS. 

Rule.  I.  Separate  the  composite  number  into  two  or  more 
factors, 

II.  Multiply  the  multiplicand  hy  one  of  these  factors,  and  that 
product  hy  another,  and  so  on  until  all  the  factors  have  been  used 
successively ;  the  last  product  will  he  the  product  required, 

KoTE. — The  factors  may  be  used  in  any  order  that  is  most  convenient,  (95,  II). 
EXAMPLES    FOR   PRACTICE. 

1.  Multiply  736  by  24.  Ans,  17664. 

2.  Multiply  538  by  56.  Ans.  30128. 

3.  Multiply  27865  by  84. 

4.  Multiply  7856  by  144.  Ans.  1131264. 

5.  What  will  56  horses  cost  at  185  each? 

6.  If  a  river  discharge  17740872  cubic  feet  of  water  in  one 
hour,  how  much  will  it  discharge  in  96  hours  ? 

Ans    1703123712  cubic  feet. 

CASE    II. 

99.   When  the  multiplier  is  a  unit  of  any  order. 

If  we  annex  a  cipher  to  the  multiplicand,  each  figure  is  removed 
one  place  toward  the  left,  and  consequently  the  value  of  the  whole 
number  is  increased  tenfold,  (57,  III).  If  two  ciphers  are 
annexed,  each  figure  is  removed  two  places  toward  the  left,  and 
the  value  of  the  number  is  increased  one  hundred  fold ;  and  every 
additional  cipher  increases  the  value  tenfold.     Hence,  the 

Rule  Annex  as  many  ciphers  to  the  multiplicand  as  there  are 
ciphers  in  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  364  by  100.  Ans.  36400. 

2.  Multiply  248  by  1000.  Ans.  248000. 

3.  What  cost  1000  head  of  cattle  at  50  dollars  each  ? 

4.  Multiply  one  million  by  one  hundred  thousand  ? 

5.  How  many  letters  will  there  be  on  100  sheets,  if  each  sheet 
have  100  Hnes,  and  each  line  100  letters  ?  Ans.  1000000. 


MULTIPLICATION. 


CASE  III. 


43 


10®,  YvHicn  there  are  ciphers  at  the  right  hand  of  one 
or  both  of  the  fiictors. 
1.  Multiply  7200  by  40. 

OPERATiox.  Analysis.     The   multiplicand,  factored,  is 

7200  equal  to  72  X  100;  the  multiplier,  factored,  is 

40  equal  to  4  X  10 ;  and  as  these  factors  taken  in 

288000  ^^^^y  order  will  give  the  same  product,  (95,  II), 

v,^e  first  multiply  72  by  4,  then  this  product 

by  100  by  annexing  two  ciphers,  and  this  product  by  10  by  annexing 

one  cipher.     Hence,  the  following 

liULE.  Maltlplij  the  slgnifivant  figures  of  the  multiplicand  hy 
those  oj^  the  7nultq^ller,  and  to  the  product  annex  as  mani/  ciphers 
as  there  are  ciphers  on  the  right  of  both  factoids. 


EXAMPLES   FOR   PRACTICE. 


1.  Multiply  740  by  800. 

2.  Multiply  36000  by  240. 

3.  Multiply  20700  by  500. 

4.  Multiply  4007000  by  3002. 
6.  Multiply  300200  by  640. 


Ans.   222000. 
Ans.   8640000. 

Ans,   12029014000. 


CASE   IV. 

101.  "When  one  part  of  the  multiplier  is  a  factor  of 
another  part. 


1. 


Multiply 

4739 

by  357. 

OPERATION. 

4739 

357 

33173 

Prod,  by  7  units. 

165865 

Prod,  by  35  tens 

1691823 

Ans. 

multiplicand  by  7  X  5  X  10 

tial  products  must  be  the  whole  product  required. 


Analysis.  In  this  example,  7,  one 
part  of  the  multiplier,  is  a  factor  of 
35,  the  other  part.  We  first  find,  in 
the  usual  manner,  the  product  of  the 
multiplicand  by  the  7  units  ;  multi- 
plying this  product  by  5,  and  writing 
the  first  figure  of  the  result  in  tens' 
place,  we  obtain  the  product  of  the 
35  tens  ;  and  the  sum  of  these  two  par' 


44  SIMPLE  NUMBERS. 

2.  Multiply  58327  by  21318. 

OPERATION.  Analysis.     In  this  exam 

rQq97  pie,  the  3  hundreds  is  a  factor 

o-j  o-j  ^  of  18,  the  part  on  the  right  of 

it,  and  also  of  21,  the  part  on 


174981         Prod,  by  3  hundreds.  ^^le  left  of  it.      We  first  mul- 

1049886    P-d.  by  18  units.  ^  ^^^    ^^.^^ 

1224867  Prod,  b^  21  thousands.         n  •      \         i      j,       i 
ngure    m    hundreds     place ; 

1243414986  Arts.  multiplying  this  product  by 

6,  and  writing  the  first  figure 
In  units'  place,  we  obtain  the  product  of  the  multiplicand  by  3  X  6  = 
18  units  ;  multiplying  the  first  partial  product  by  7,  and  writing  the 
first  figure  in  thousands'  place,  we  obtain  the  product  of  the  multipli- 
cand by  7  X  3  X  1000  =  21  thousands ,  and  the  sum  of  these  three 
partial  products  must  be  the  entire  product  required. 

Note. — The  product  obtained  by  multiplying  any  partial  product  is  called  a 
derived  product. 

103*    From  these  illustrations  we  have  the  following 

KuLE.     I.  Find  the  product  of  the  multiplicand  hi/  some  figure 

of  the  multiplier  which  is  a  factor  of  one  or  more  parts  of  the 

midtiplier. 

II.  Multiply  this  product  hy  that  factor  ichich,  taken  with  the 
figure  of  the  multiplier  first  used,  icill  produce  other  pa7'ts  of  the 
multiplier  J  and  write  the  first  figure  of  each  residt  under  the  first 
figure  of  the  part  of  the  midtiplier  thus  used. 

III.  In  like  manner,  find  the  product,  either  direct  or  derived, 
for  every  figure  or  part  of  the  multiplier ;  the  sum  of  all  the  pro- 
ducts will  he  the  whole  product  required. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  5784  by  246.  Ans.  1422864. 

2.  Multiply  3785  by  721.  Ans.  2728985. 

3.  Multiply  472856  by  54918.  Aiis.  25968805808. 

4.  Multiply  43785  by  7153.  Ans.  313194105. 

5.  Multiply  573042  by  24816.  Ans.  14220610272. 

6.  Multiply  78563721  by  127369. 

7.  Multiply  43r25652  by  5187914. 


MULTIPLICATION.  45 

8.  Multiply  3578426785  by  64532164. 

9.  Multiply  2703605  by  4249784. 

10.  What  is  the  product  of  9462108  multiplied  by  16824? 

Ans,  159,190,504,992. 

EXAMPLES   COMBINING   THE   PRECEDING    RULES. 

1.  A  man  bought  two  farms,  one  containing  175  acres  at  $28 
per  acre,  and  the  other  containing  320  acres  at  $37   per  acre 
what  was  the  cost  of  both  ?  Ans.  $16,740. 

2.  If  a  man  receive  $1200  salary,  and  pay  $364  for  board 
$275  for  clothing,  $150  for  books,  and  $187  for  other  expenses, 
how  much  can  he  save  in  5  years?  Ans.  $1,120. 

3.  Two  persons  start  from  the  same  point,  and  travel  in  oppo 
site  directions;  one  travels  29  miles  a  day,  and  the  other  32  miles, 
How  far  apart  will  they  be  in  17  days  ?  Ans.  1,037  miles. 

4.  A  drover  bought  127  head  of  cattle  at  $34  a  head,  and  97 
head  at  $47  a  head,  and  sold  the  whole  lot  at  $40  a  head ;  what 
was  his  entire  profit  or  loss  ?  Ans.  $83  profit. 

5.  Multiply  675  — (77  +  56)  by  (3  x  156)  — (214  — 28). 

Ans.  152844. 

6.  Multiply  98  4  6  x  (37  +  50)  by  (64  —  50)  x  5  — 10. 

Ans.  37200. 

7.  What  is  the  product  of  (14  X  25)  —  (9'x  36)  +  4324  x 
(280  — 112)  +  (376"  +  42)  X  4  ?  Ans.  8,004,000. 

8.  In  1850  South  Carolina  cultivated  29967  farms  and  planta- 
tions, containing  an  average  of  541  acres  each,  at  an  average  value 
of  $2751  for  each  farm;  New  Jersey  cultivated  23905  farms,  con- 
taining an  average  of  115  acres  each,  at  an  average  value  of 
$5030  per  farm.  How  much  more  were  the  farming  lands  of  the 
latter  valued  at,  than  those  of  the  former  ? 

9.  There  are  in  the  United  States  1922890880  acres  of  land ; 
of  this  there  were  reported  under  cultivation,  in  1850,  1449075 
farms,  each  embracing  an  average  of  203^  acres.  How  many  acres 
were  still  uncultivated  ? 

10.  Each  of  the  above  farms  in  the  United  States  was  valued 
at  an  average  of  $2258,  and  upon  each  farm  there  was  an  average 


46  SIMPLE  NUMBERS. 

of  $105  in  implements  and  machinery.    What,  was  the  aggregate 
value  of  the  farms  and  implements?  Ans.  $3,424,164,225. 

Find  the  values  of  the  following  expressions  : 

11.  2*x  5^  —  7^?  Ans.  49,657. 

12.  15^  — (3-^  X  2^)  -f  208^  —  9  x  2*?  Ans.  46,207. 

13.  2^  +  3^  -f  4^  +  5^  4  6«  ? 

14.  In  1852  Great  Britain  consumed  1200000  bales  of  American 
cotton ;  allowing  each  bale  to  contain  400  pounds,  what  was  its 
total  weight? 

15.  If  a  house  is  worth  §2450,  and  the  farm  on  which  it  stands 
6  times  as  much,  lacking  $500,  and  the  stock  on  the  farm  twice 
as  much  as  the  house,  what  is  the  value  of  the  whole  ? 

Ans.  $21550. 

16.  A  flour  merchant  bought  1500  barrels  of  flour  at  7  dollars 
a  barrel ;  he  sold  800  barrels  at  10  dollars  a  barrel,  and  the  re- 
mainder at  6  dollars  a  barrel.     How  much  was  his  gain  ? 

17.  A  man  invests  in  trade  at  one  time  $450,  at  another  $780, 
at  another  $1250,  and  at  another  $2275 ;  how  much  must  he  add 
to  these  sums,  that  the  amount  invested  by  him  shall  be  increased 
fourfold?  Ans.  $14,265. 

18.  At  the  commencement  of  the  year  1858  there  were  in  ope- 
ration in  the  United  States  35000  miles  of  telegraph;  allowing 
the  average  cost  to  be  $115  per  mile,  what  was  the  total  cost? 

19.  The  cost  of  the  Atlantic  Telegraph  Cable,  as  originally 
made,  was  as  follows;  2500  miles  at  $485  per  mile,  10  miles  deep- 
sea  cable  at  $1450  per  mile,  and  25  miles  shore  ends  at  $1250  per 
mile.     What  was  its  total  cost?  Ans.  $1,258,250. 

20.  For  the  year  ending  June  SO,  1859,  there  were  coined  in 
the  United  States  1401944  double  eagles  valued  at  twenty  dollars 
each,  62990  eagles,  154555  half  eagles,  and  22059  three  dollar 
pieces ;  what  was  the  total  value  of  this  gold  coin  ? 

Ans,  $29;507,732. 


DIVISION.  4*^ 


DIVISION. 

i03*  Division  is  the  process  of  finding  how  many  times  one 
number  is  contained  in  another. 

104,    The  Dividend  is  the  number  to  be  divided. 

10^.    The  Divisor  is  the  number  to  divide  by. 

l&Q.  The  Quotient  is  the  result  obtained  by  the  process  of 
division. 

|®y.    The  Reciprocal  of  a  number  is  1  divided  by  the  number. 

Tims,  the  reciprocal  of  15  is  1  -^  15,  or  y*g. 

NoTKS. — 1.  When  the  dividend  does  not  contain  the  divisor  an  exact  number 
of  times,  the  part  of  the  dividend  left  is  called  the  liemainder,  which  must  bo 
less  than  the  divisor. 

2.  As  the  remainder  is  always  a  part  of  the  dividend,  it  is  always  of  the  same 
name  or  kind.  ' 

3.  When  there  is  no  remainder  the  division  is  said  to  be  exact. 

108.  The  method  of  dividing  any  number  by  another  depends 
upon  the  following  principles : 

I.  Division  is  the  reverse  of  multiplication,  the  dividend  cor- 
responding to  the  product,  and  the  divisor  and  quotient  to  the 
factors. 

II.  If  all  the  parts  of  a  number  be  divided,  the  entire  number 
will  be  divided. 

Since  the  remainder  in  dividing  any  part  of  the  dividend  must 
be  less  than  the  divisor,  it  can  be  divided  only  by  being  expressed 
m  units  of  a  lower  order.     Hence, 

III.  The  operation  must  commence  with  the  units  of  the  high- 
est order. 

1.  Divide  2742  by  6. 

Analysis.     We  write  the  divisor  at  the  Itft 

OPERATIOX.  n     1        -,.    .1        -,  in  •      T  T 

01  the  dividend,  separated  from  it  by  a  line. 

^  Z "1  As  6  is  not  contained  in  2  thousands,  we  take 

457  Ans.  the  2  thousands  and  7  hundreds  together,  and 

proceed  thus ;  6  is  contained  in  27  hundreds 

4  hundred  times,  and  the  remainder  is  3  hundreds ;  we  write  4  in 

hundreds^  place  in  the  quotient,  and  unite  the  remainder,  3  hundreds, 


48  SIMPLE  NUMBERS. 

to  the  next  figure  of  the  dividend,  making  34  tens ;  then,  6  is  con- 
tained in  34  tens  5  tens  times,  and  the  remainder  is  4  tens ;  writing 
5  tens  in  its  place  in  the  quotient,  we  unite  the  remainder  to  the  next 
figure  in  the  dividend,  making  42 ;  6  is  contained  in  42  units  7  times, 
and  there  is  no  remainder ;  writing  7  in  its  place  in  the  quotient,  we 
have  the  entire  quotient,  457. 

Note  1. — The  dififerent  numbers  which  we  divide  in  obtaining  the  successive 
figures  of  the  quotient,  are  called  partial  dividends. 


2.  Divide  18149  by  56. 

OPERATIOX. 


Analysis.     As  neither  1  nor  18 
will  contain  the  divisor,  we  take 
66 )  18149  (324/g  Ans.  ^^ree  figures,  181,  for  the  first  par- 

^  tial  dividend.     56  is  contained  in 

134  181  3  times,  and  a  remainder ;  we 

112  write  the  3  as  the  first  figure  in 

229  the  quotient,  and  then   multiply 

224  the  divisor  by  this  quotient  figure ; 

^  3  times  56  is  168,  which  subtracted 

from  181,  leaves  13 ;  to  this  re- 
mainder we  annex  or  bring  down  4,  the  next  figure  of  the  dividend, 
and  thus  form  134,  the  next  partial  dividend ;  56  is  contained  in 
134  2  times,  and  a  remainder;  2  times  56  is  112,  which  subtracted 
from  134,  leaves  22;  to  this  remainder  we  bring  down  9,  the  last 
figure  of  the  dividend,  and  we  have  229,  the  last  partial  dividend ;  56 
is  contained  in  229  4  times,  and  a  remainder ;  4  times  56  is  224, 
which  subtracted  from  229,  gives  5,  the  final  remainder,  which  we 
write  in  the  quotient  with  the  divisor  below  it,  thus  completing  tho 
division,  (35). 

Note  2. — When  the  multiplication  and  subtraction  are  performed  mentally, 
as  in  the  first  example,  the  operation  is  called  Short  Diviaiov ;  but  when  the 
work  is  written  out  in  full,  as  in  the  second  example,  the  operation  is  called 
Loti(/  Division.     The  principles  governing  the  two  methods  are  the  same. 

109*  From  these  principles  and  ilbistrations  we  derive  the 
following  general 

Rule.  I.  Beginning  at  the  left  liandj  tahe  for  the  fir ^t  partial 
dividend  the  fewest  figures  of  the  given  dividend  that  will  contain 
the  divisor  one  or  more  times  ;  find  how  many  times  the  divisor  is 
contained  in  this  partial  dividend,  and  icrite  the  result  in  the 
quotient;  multiply  the  divisor  hy  this  quotient  figure^  and  subtract 
the  product  from  the  partial  dividend  used. 


DIVISION. 


49 


II.  To  the  remainder  hring  down  the  next  figure  of  the  dividend^ 
with  which  proceed  as  he/ore;  and  thus  continue  till  all  the  figures 
of  the  dividend  have  been  divided. 

III.  If  the  division  is  not  exact,  place  the  final  remainder  in 
the  quotient,  and.  write  the  divisor  underneath. 

no.  Proof.  There  are  two  principal  methods  of  proving 
division. 

1st.  By  multiplication. 

Multiply  the  divisor  and  quotient  together,  and  to  the  product 

iidd  the  remainder,  if  any;  if  the  result  be  equal  to  the  dividend, 

the  work  is  correct.     (108,  I.) 

Note. — In  multiplication,  the  two  factors  are  given  to  find  the  product;  in 
division,  the  product  and  one  of  the  factors  ar«  given  to  find  the  other  factor. 

2d.  By  excess  of  9's. 

11  fl.  Subtract  the  remainder,  if  any,  from  the  dividend,  and 
find  the  excess  of  9's  in  the  result.  Multiply  the  excess  of  9's  in 
the  divisor  by  the  excess  of  9's  in  the  quotient,  and  find  the  excess 
of  9's  in  the  product;  if  the  latter  excess  is  the  same  as  the 
former,  the  work  is  supposed  to  be  correct.    (8o.) 

EXAMPLES   FOR  PRACTICE. 


(1.)                   (2.)                   (3.) 

(4.) 

6)473832        8)972496       9)1370961 

12)73042164 

Quotients. 

5.  Divide  170352  by  36. 

4732. 

6.  Divide  409887  by  47. 

8721. 

7.  Divide  443520  by  84. 

5280. 

8.  Divide  36380250  by  125. 

291042. 

9.  Divide  1554768  by  216. 

10.  Divide  3931476  by  556. 

11.  Divide  48288058  by  3094. 

Rem. 

12.  Divide  11214887  by  232. 

7. 

13.  Divide  27085946  by  216. 

194. 

14.  Divide  29137062  by  5317. 

5219. 

15.  Divide  4917968967  by  2359. 

1255. 

5                                                    D 

60  SIMPLE  NUMBERS. 

16.  What  is  the  value  of  721198  ~  291  ?  Rem.  100. 

17.  What  is  the  value  of  3844449  -^  657  ?  342. 

18.  What  is  the  value  of  536819237  -^  907  ?  403. 

19.  What  is  the  value  of  571943007145  -^  37149  ?      12214. 

20.  What  is  the  value  of  48659910--  54001  ?  5009. 

21.  The  annual  receipts  of  a  manufacturing  company  are 
$147675;  how  much  is  that  per  day,  there  being  365  da3^s  in  the 
year?  Ans.  $404|J|. 

22.  The  New  York  Central  Kailroad  Company,  in  1 859,  owned 
556  miles  in  length  of  railroad,  which  cost,  for  construction  and 
equipment,  $30732518;  what  was  the  average  cost  per  mile? 

J«.5.  §55,2741-11. 

23.  The  Memphis  and  Charleston  Railroad  is  287  miles  in 
length,  and  cost  $5572470 ;  what  was  the  average  cost  per  mile  ? 

Ans.  $19,4162^3^-. 

24.  The  whole  number  of  Post  offices  in  the  United  States,  in 
1858,  was  27977,  and  the  revenue  was  $8186793 ;  what  was  the 
average  income  to  an  office  ? 

ABBREVIATED   LONG   DIVISION. 
liSo    We  may  avoid  writing  the  products  in  long  division, 
and  obtain  the  successive  remainders  by  the  method  of  subtraction 
employed  in  the  case  of  several  subtrahends,     (yij,) 

1.  Divide  261249  by  487. 

OPERATiox.  Analysis.     Dividing  the  first  partial 

487 ")  261249  (  536  dividend,  2612,  we  obtain  5  for  the  first 

177  figure  of  the  quotient.     We  now  multi- 

313  ply  487  by  5  ;  but  instead  of  writing  the 

217  Rem.  product,  and    subtracting   it  from  the 

partial    dividend,   we    simply  observe 

what  figures  must  be  added  to  the  figures  of  the  product,  as  we  proceed, 

to  give  the  figures  of  the  partial  dividend,  and  write  them  for  the 

remainder  sought.     Thus,  5  times  7  are  35,  and  7  (written  in  the 

remainder,)  are  42,  a  number  whose  unit  figure  is  the  same  as  the 

right  hand  figure  of  the  partial  dividend  ;  5  times  8  are  40,  and  4,  the 

tens  of  the  42,  are  44,  and  7  (written  in  the  remainder,)  are  51 ;  5 


DIVISION. 


61 


times  4  are  20,  and  5,  the  tens  of  the  51,  are  25,  and  1  (written  in  the 
remainder,)  are  26.  We  next  consider  the  whole  remainder,  177,  as 
joined  with  4,  the  next  figure  of  the  dividend,  making  1774  for  the 
next  partial  dividend.  Proceeding  as  before,  we  obtain  313  for  the 
second  remainder,  217  for  the  final  remainder,  and  53G  for  the  entire 
quotient.     Hence,  the  following 

Rule.  I.  Obtain  the  first  figure  in  the  quotient  in  the  usual 
manner. 

II.  Multiply  the  first  figure  of  the  divisor  hy  this  quotient  figure, 
and  write  such  a  figure  in  the  remainder  as,  added  to  this  partial 
product,  will  give  an  amount  having  for  its  unit  figure  the  first  or 
right  hand  figure  of  the  partial  dividend  used. 

III.  Carry  the  tens^  figure  of  the  amount  to  the  product  of  the 
next  figure  of  the  divisor,  and  proceed  as  before,  till  the  entire 
remainder  is  obtained. 

TV.  Conceive  this  remainder  to  be  joined  to  the  next  figure  of 
the  dividend,  for  a  new  partial  dividend,  and  proceed  as  with  the 
former,  till  the  work  is  finished, 

EXAMPLES   FOR   PRACTICE. 

1.  Divide  77112  by  204.  Ans.  378. 

2.  Divide  65664  by  72.  Ans.  912. 

3.  Divide  7918576  by  209.  Ans.  37864. 

4.  Divide  6636584  by  698. 

5.  Divide  4024156  by  8903.  Ans.  452. 
.     6.  Divide  760592  by  6791. 

7.  Divide  101443929  by  25203.  Ans.  4.025^\%%. 

8.  Divide  1246038849  by  269181.  Ans.  4629. 

9.  Divide  2318922  by  56240. 

10.  Divide  1454900  by  17300.  Ans.  S4j\\%%. 

GENERAL  PRINCIPLES  OF  DIVISION. 
ll*l.    The  general  principles  of  division  most  important  in  their 
application,  relate;    1st,   to   changing  the   terms  of  division   by 
addition  or  subtraction;    2d,  to  changing  the  terms  of  division 
by  multiplication  or  division ;  3d,  to  successive  division. 


52  SIMPLE  NUMBERS. 

114:t  The  quotient  in  division  depends  upon  the  relative  values 
of  the  dividend  and  divisor.  Hence,  any  change  in  the  value  of 
either  dividend  or  divisor  must  produce  a  change  in  the  value  of 
the  quotient;  though  certain  changes  may  be  made  in  both  divi- 
dend and  divisor^  at  the  same  time,  that  will  not  affect  the  quotient. 

CHANGING   THE   TERMS   BY   ADDITION   OR   SUBTRACTION. 

11^.  Since  the  dividend  corresponds  to  a  product,  of  which 
the  divisor  and  quotient  are  factors,  we  observe, 

1st.  If  the  divisor  be  increased  by  1,  the  dividend  must  be 
increased  by  as  many  units  as  there  are  in  the  quotient,  in  order 
that  the  quotient  may  remain  the  same,  (03,  I) ;  and  if  the  divi- 
dend be  Twt  thus  increased,  the  quotient  will  be  diminished  by  as 
many  units  as  the  number  of  times  the  new  divisor  is  contained 
in  the  quotient.     Thus, 

84  -f-  6  =  14 

84  --  7  =  14  —  V  =  12 

2d.  If  the  divisor  be  diminished  by  1,  the  dividend  must  be 
diminished  by  as  many  units  as  there  are  in  the  quotient,  in  order 
that  the  quotient  may  remain  the  same,  (03,  II) ;  and  if  the 
dividend  be  not  thus  diminished,  the  quotient  will  be  increased 
by  as  many  units  as  the  number  of  times  the  new  divisor  is  con- 
tained in  the  quotient.     Thus, 

144  -f-  9  ==  16 

144  --  8  =  16  +  \^  =  18 

These  principles  may  be  stated  as  follows : 

I.  Addiruj  1  to  the  divisor  takes  as  mani/  units  from  the  quotient 
as  the  new  divisor  is  contained  times  in  the  quotient. 

II.  Subtracting  1  from  the  divisor  adds  as  many  units  to  the 
quotient  as  the  new  divisor  is  contained  times  in  the  quotient. 
Htnce, 

III.  Adding  any  number  to  the  divisor  subtracts  as  many  units 
from  the  quotient  as  the  new  divisor  is  contained  times  in  the  pro- 
duct of  the  quotient  by  the  number  added;   and  subtracting 


I 


DIVISION. 


63 


r 


fny  nwnher  from  the  divisor  ADDS  as  many  units  to  the  quotient 
as  the  new  divisor  is  contained  times  in  the  product  of  the  quo- 
lent  by  the  number  subtracted. 


CHANGING   THE   TERMS    BY    MULTIPLICATION    OR   DIVISION. 

116.    There  are  six  cases : 

1st.  If  any  divisor  is  contained  in  a  given  dividend  a  certain 
number  of  times,  the  same  divisor  will  be  contained  in  twice  the 
dividend  twice  as  many  times;  in  three  times  tjie  dividend,  three 
times  as  many  times;  and  so  on.     Hence, 

MuUiplyiug  the  dividend  by  any  nwnbeVj  multiplies  the  quotient 
by  the  same  number. 

2d.  If  any  divisor  is  contained  in  a  given  dividend  a  certain 
number  of  times,  the  same  divisor  will  be  contained  in  one  half 
the  dividend  one  half  as  many  times )  in  one  third  the  dividend, 
one  third  as  many  times;  and  so  on.     Hence, 

Dividing  the  dividend  by  any  number ^  divides  the  quotient  by  the 
same  number. 

8d.  If  a  given  divisor  is  contained  in  any  dividend  a  certain 
number  of  times,  twice  the  divisor  will  be  contained  in  the  same 
dividend  one  half  as  many  times ;  three  times  the  divisor,  one 
third  as  many  times ;  and  so  on.     Hence, 

Multiplying  the  divisor  by  any  number j  divides  the  quotient  by 
the  same  number. 

4th.  If  a  given  divisor  is  contained  in  any  dividend  a  certain 
number  of  times,  one  half  the  divisor  will  be  contained  in  the 
same  dividend  twice  as  many  times ;  one  third  of  the  divisor,  three 
times  as  many  times;  and  so  on.     Hence, 

Dividing  the  divisor  by  any  number^  multip)lies  the  quotient  by 
the  same  number. 

5th.  It  a  given  divisor  is  contained  in  a  given  dividend  a  cer- 
tain number  of  times,  twice  the  divisor  will  be  contained  the  same 
number  of  times  in  twice  the  dividend;  three  times  the  divisor 
will  be  contained  the  same  number  of  times  in  three  times  the 
dividend ;  and  so  on.  Hence, 
5* 


54  SIMPLE  NUMBERS. 

Multiplying  both  dividend  and  divisor  hy  the  same  number  does 
not  alter  the  quotient. 

6th.  if  a  given  divisor  is  contained  in  a  given  dividend  a  cer^ 
tain  number  of  times,  one  half  the  divisor  will  be  contained  the 
same  number  of  times  in  one  half  the  dividend;  one  third  of  the 
divisor  will  be  contained  the  same  number  of  times  in  one  third 
of  the  dividend ;  and  so  on.     Hence, 

Dividing  both  dividend  and  divisor  by  the  same  number  does  not 
alter  the  quotient. 

Note. — If  a  number  be  multiplied  and  the  product  divided  by  the  same  num- 
ber, the  quotient  will  be  equal  to  the  number  multiplied;  hence  the  5th  case 
may  be  regarded  as  a  direct  consequence  of  the  1st  and  3d;  and  the  6th,  as  the 
direct  consequence  of  the  2d  and  4th. 

To  illustrate  these  cases,  take  24  for  a  dividend  and  6  for  a 
divisor ;  then  the  quotient  will  be  4,  and  the  several  changes  may 
be  represented  in  theii  order  as  follows : 


1      4.9    _i_    fi    ft  I  Multiplying  the  dividend  by  2  multi- 

^       plies  the  quotient  by  2. 

o     io_:.p     of  Dividing   the   dividend  by  2  divides 

~  the  quotient  by  2. 

Multiplying  the  divisor  by  2  divides 
the  quotient  by  2. 

Dividing   the  divisor  by  2   multiplies 
the  quotient  by  2. 

c      4Q     _i_i9     A  [  Multiplying  both  dividend  and  divisor 

I      by  2  does  not  alter  the  quotient. 

/>      19    _i_    Q     j^  (  Dividing  both  dividend  and  divisor  by 

1      2  does  not  alter  the  quotient. 

11^.  These  six  cases  constitute  three  general  principles, 
which  may  now  be  stated  as  follows : 

Prin.  I.  Multiplying  the  dividend  multiplies  the  quotient)  and 
dividing  the  dividend  divides  the  quotient. 

Prin.  II.  Multiplying  the  divisor  divides  the  quotient;  and 
dividing  the  divisor  midtiplies  the  quotient. 


Dividend. 

24 

Divisor 

--    6 

.      Quotient. 

=    4 

.    48 

--    6 

=  'r 

.    12 

-H    6 

=  .{: 

:.    24 

-v-12 

=  .{: 

.    24 

--    3 

=  s{: 

b 


DIVISION. 


55 


Prin.  III.     Multiplying  or  dividing  both  dividend  and  divisor 
hy  the  same  numher,  does  not  alter  the  quotient. 

118»    These  three  principles  may  be  embraced  in  one 


GENERAL    LAW. 

A  change  in  the  dividend  produces  a  like  change  in  the  quo- 
tient;  hut  a  change  in  the  divisor  produces  an  OPPOSITE  change 
in  the  quotient. 

SUCCESSIVE  DIVISION. 

119.  Successive  Division  is  the  process  of  dividing  one 
number  by  another,  and  the  resulting  quotient  by  a  second  divisor, 
and  so  on. 

Successive  division  is  the  reverse  of  continued  multiplication. 
Hence, 

I.  If  a  given  number  be  divided  by  several  numbers  in  succes- 
sive division,  the  result  will  be  the  same  as  if  the  given  number 
were  divided  by  the  product  of  the  several  divisors,  (9S,  I). 

II.  The  result  of  successive  division  is  the  same,  in  whatever 
order  the  divisors  are  taken,  (9^,  II). 

• 

CONTRACTIONS  IN  DIVISION. 
CASE    I. 

120.  When  the  divisor  is  a  composite  number. 
1.  Divide  1242  by  54. 

Analysis.     The  component  factors  of  54  are 

OPERATION.  ,  ^ 

6  and  9.     We  divide  1242  by  6,  and  the  rc- 

^    ^  suiting  quotient  by  9,  and  obtain  for  the  final 

9)207  result,  23,  which  must  be  the  same  as  the 

23  Ans.         quotient  of  1242  divided  by  0  times  9,  or  54, 

(119,  I).     We  might  have  obtained  the  same 

result  by  dividing  first  by  9,  and  then  by  G,  (119,  II).      Hence  the 

following 

KuLE.     Divide  the  dividend  hy  one  of  the  factors^  and  the  quos 


66  SIMPLE  NUMBERS. 

tient  thus  obtained  hy  another ,  and  so  on  if  there  he  more  than  two 
factors,  until  every  factor  has  been  made  a  divisor.  The  last  quo- 
tient will  be  the  quotient  required,^ 


TO   FIND   THE   TRUE   REMAINDER. 

131.  If  remainders  occur  in  successive  division,  it  is  evident 
that  the  true  remainder  must  be  the  least  number,  which,  sub- 
tracted from  the  given  dividend,  will  render  all  the  divisions 
exact 

I.  Divide  5855  by  168,  using  the  factors  3,  7,  and  8,  and  find 
the  true  remainder, 

OPERATION.  Analysis.      Dividing    the 

S)  5855  given  dividend  by  3,  we  have 

7  "i  1951  2  "^^^  ^^^  ^  quotient,   and  a 

remainder   of  2.     Hence,   2 

^)^ 5x3=    15  subtracted  from  5855  would 

34  ...6x  7x  3=1 26  render  the  first  division  exact, 

True  remainder 143  and  we  therefore  write  2  for 

a  part  of  the  true  remainder. 
Dividing  1951  by  7,  we  have  278  for  a  quotient,  and  a  remainder  of  5. 
Hence,  5  subtracted  from  1951  would  render  the  second  division  exact. 
But  to  diminish  1951*by  5  would  require  us  to  diminish  1951  X  3,  the 
divide*id  of  the  first  exact  division,  by  5  X  3  —  15,  (93,  III) ;  and 
we  therefore  write  15  for  the  second  part  of  the  true  remainder. 
Dividing  278  by  8,  we  have  34  for  a  quotient,  and  a  remainder  of  6. 
Hence,  6  subtracted  from  278  would  render  the  third  division  exact. 
But  to  diminish  278  by  6  would  require  us  to  diminish  278  X  7,  the 
dividend  of  the  second  exact  division,  by  6  X  7 ;  or  278  X  7  X  3,  the 
dividend  of  the  first  exact  division,  by  6x7x3  =  126 ;  and  we 
therefore  write  126  for  the  third  part  of  the  true  remainder.  Adding 
the  three  parts,  we  have  143  for  the  entire  remainder. 

Hence  the  following 

Rule.  L  Multiph/  each  j)artial  remainder  hy  all  the  preceding 
divisors. 

II.  Add  the  several  products;  the  sum  will  he  the  true  re^ 
mainder. 


DIVISION. 


57 


1.  D 

2.  D 

3.  D 

4.  D 

5.  D 

6.  D 

7.  D 

8.  D 

9.  D 

10.  D 

11.  D 

12.  D 

13.  D 

14.  D 

15.  D 

16.  D 

17.  B 

18.  Di 

19.  D 


EXAMPLES    FOR   PRACTICE. 

ide  435  by  15  =  3  X  5.  Ans.   29. 
de  425G  by  56  =  7  X  8. 
de  17856  by  72  =  9  x  8. 

[de  15288  by  42  :==  2  x  3  x  7.  Ans.   364. 

[de  972552  by  168  =  8  x  7  X  3.  Ans,   5789. 
[de  526050  by  126  =  9  X  7  X  2. 

[de  612360  by  105  =  7  x  5  x  3.  Ans.   5832. 

ide  553  by  15  =  3  X  5.  Rem.  13. 

de  10183  by  105  =  3  X  5  X  7.  103. 

de  10197  by  120  =  2  X  3  X  4  X  5.  117. 

de  29792  by  144  =  3  x  8  x  6.  128. 

de  73522  by  168  =  4  x  6  x  7.  106. 

de  63814  by  135  =  3  X  5  X  9.  ^    124. 
[de  386639  by  720  =  2  x  3  x  4  x  5  x  6.'    719. 

de  734514  by  168  =  4  x  6  x  7.  18. 

[de  636388  by  729  =  9^.  700. 

de  4619  by  125  =  5^  119. 
de  116423  by  10584  =  3  x  7^  x  8  x  9.    10583. 

de  79500  by  6125  =  5^  x  T.  6000. 


CASE   II. 

1$53.   "When  the  divisor  is  a  unit  of  any  order. 

If  we  cut  off  or  remove  the  right  hand  figure  of  a  number,  each 
of  the  other  figures  is  removed  one  place  toward  the  right,  and, 
consequently,  the  value  of  each  is  diminished  tenfold,  or  divided 
by  10,  (37^  III).  For  a  similar  reason,  by  cutting  off  two  figures 
we  divide  by  100 ;  by  cutting  off  threcj  we  divide  by  1000,  and 
BO  on ;  and  the  figures  cut  off  will  constitute  the  remainder. 
Hence  the  ^mj^ 

Rule.  Ftotyi  the  right  hand  of  the  dividend  cut  off  as  many 
figures  as  there  are  ciphers  in  the  divisor.  Under  the  figures  so 
cut  off,  place  the  divisor,  and  the  whole  will  form  the  quotienL 


58  SIMPLE  NUMBERS. 

EXAMPLES   FOR   PRACTICE. 

1.  Divide  79  by  10.  Ans.  7-i%. 

2.  Divide  7982  by  100. 

3.  Divide  4003  by  1000.  Ans.  4y^3^^. 

4.  Divide  2301050  by  10000. 

6.  Divide  3600036  by  1000.  Ans.  3600tMtj- 

CASE   III. 

133.   When  there  are  ciphers  on  the  right  hand  of 
the  divisor. 

I.  Divide  25548  by  700. 

Analysis.     "We  resolve  700 

OPERATION.  .  ,        ^  -.^^         -,    ^ 

into   the   factors    100   and  7. 

7|00)  255148  Dividing  first  by  100,  the  quo- 

36  Quotient.   3  2d  rem.  tient  is  255,  and  the  remainder 

3  X  100  +  48  =  348  true  rem.       48.    Dividing  255  by  7,  the 

final  quotient  is  36,  and  the 

second  remainder  3.      Multiplying   the   last   remainder,   3,  by  the 

preceding  divisor,  100,  and  adding  the  preceding  remainder,  we  have 

300 -f  48  =  348,  the  true  remainder,  (121).     In  practice,  the  true 

remainder  may  be  obtained  by  prefixing  the  second  remainder  to 

the  first.     Hence  the 

HuLE.     I.    Cut  off  the  ciphers  from  the  right  of  the  divisor ,  and 

as  many  figures  from  the  right  of  the  dividend. 

II.  Divide  the  remaining  figures  of  the  dividend  hy  the  remain- 
ing  figures  of  the  divisor j  for  the  final  quotient. 

III.  Prefix  the  remainder  to  the  figures  cut  off^  and  the  result 
will  he  the  true  remainder. 

EXAMPLES   FOR   PRACTICE. 

1.  Divide  7856  by  900.  Ans.  8f  §f 

2.  Divide  13872  by  500. 

3.  Divide  8||W:8  by  2600.  Ans.  ^2^^^^, 

4.  Divide  1548036  by  4300.  Ans.  ^^^^lU- 

5.  Divide  436000  by  300.  Ans.  1453|g§. 

6.  Divide  66472000  by  8100. 
r.  Divide  10S18000  by  3600. 


DIVISION. 


EXAMPLES    COMBINING   THE    TRECEDING    RULES. 


59 


1.  How  many  barrels  of  flour  at  S8  a  barrel,  will  pay  for  25  tons 
of  coal  at  $4  a  ton,  and  3G  cords  of  wood  at  §3  a  cord  ? 

Ans.  26. 

2.  A  grocer  bought  12  barrels  of  sugar  at  §16  per  barrel,  and 
17  barrels  at  $13  per  barrel ;  how  much  would  he  gain  by  selling 
the  whole  at  $18  per  barrel  ? 

3.  x\  farmer  sold  300  bushels  of  wheat  at  $2  a  bushel,  corn  and 
oats  to  the  amount  of  $750 ;  with  the  proceeds  he  bought  120 
head  of  sheep  at  $3  a  head,  one  pair  of  oxen  for  $90,  and  25  acres 
of  land  for  the  remainder  How  much  did  the  land  cost  him  per 
acre? Ans.  $36. 

4.  Divide  450+ (24  — 12)  x  5  by  (90  -f-  6)  +  (3  x  llJ^irU.. 

Ans.   17. 

5.  Divide  648  x  (3^  x  '2')  —  9  —  (2910  ~-  15)  by  2863  -^ 
(4375  -nj5)  X  4^+  3^  Ans.  712f . 

6.  The  product  of  three  numbers  is  107100;  one  of  the 
numbers  is  42,  and  another  34.     What  is  the  third  number  ? 

Ans.  7d. 

7.  What  number  is  that  which  being  divided  by  45,  the  quo- 
tient increased  by  7^  +  1,  the  sum  diminished  by  the  difference 
between  28  and  16,  the  remainder  multiplied  by  6,  and  the  pro- 
duct divided  by  24,  the  quotient  will  be  12  ?  Ans.  450. 

8.  A  mechanic  earns  $60  a  month,  but  his  necessary  expenses 
are  $42  a  month.  How  long  will  it  take  him  to  pay  for  a  farm 
of  50  acres  worth  $36  an  acre  ? 

9.  What  number  besides  472  will  divide  251104  without  a  re~ 
mainder?  Ans.  532. 

10.  Of  what  number  is  3042  both  divisor  and  quotient  ? 

Ans.  9253764. 

11.  What  must  the  number  be  which,  divided  by  453,  will  give 
the  quotient  307,  and  the  remainder  109  ?  Ans.  139180. 

12.  A  farmer  bought  a  lot  of  sheep  and  hogs,  of  each  an  equal 
number,  for  $1276.     He  gave  $4  a  head  for  the  sheep,  and  $7  a 


go  SIMPLE  NUMBERS. 

head  for  the  hogs ;  what  was  the  whole  number  purchased,  and 
how  much  was  the  difference  in  the  total  cost  of  each  ? 

Ans.  232  purchased ;  $348  difference  in  cost. 

13.  According  to  the  census  of  1850  the  total  value  of  the 
tobacco  raised  in  the  United  States  was  $13,982,686.  How  many 
school-houses  at  a  cost  of  $950,  and  churches  at  a  cost  of  $7500, 
of  each  an  equal  number,  could  be  built  with  the  proceeds  of  the 
tobacco  crop  of  1850  ?       Ans.  1654,  and  a  remainder  of  $6386. 

14.  The  entire  cotton  crop  in  the  United  States  in  1859  was 
4,300,000  bales,  valued  at  $54  per  bale.  If  the  entire  proceeds 
were  exchanged  for  English  iron,  at  $60  per  ton,  how  many  tons 
would  be  received  ? 

15.  The  population  of  the  United  States  in  1850  was  23,191,876. 
It  was  estimated  that  1  person  in  every  400  died  of  intemperance. 
How  many  deaths  may  be  attributed  to  this  cause  in  the  United 
States,  during  that  year  ? 

16.  In  1850-,  there  were  in  the  State  ot  New  York,  10,593 
public  schools,  which  were  attended  during  the  winter  by  508464 
pupils ;  what  was  the  average  number  to  each  school  ? 

Ans.  48. 

17.  A  drover  bought  a  certain  number  of  cattle  for  $9800,  and 
sold  a  certain  number  of  them  for  $7680,  at  $64  a  head,  and 
gained  on  those  he  sold  $960.  How  much  did  he  gain  a  head, 
and  how  many  did  he  buy  at  first  ? 

Ans.  Gained  $8  per  head;  bought  175. 

18.  A  house  and  lot  valued  at  $1200,  and  6  horses  at  $95  each, 
were  exchanged  for  30  acres  of  land.  At  how  much  was  the  land 
valued  per  acre  ? 

19.  If  16  men  can  perform  a  job  of  work  in  36  days,  in  how 
many  days  can  they  perform  the  same  job  with  the  assistance  of 
8  more  men  ?  Ans.  24. 

.  20.  Bought  275  barrels  of  flour  for  $1650,  and  sold  186  bar- 
rels of  it  at  $9  a  barrel,  and  the  remainder  for  what  it  cost.  How 
much  was  gained  by  the  bargain  ?  Ans.  $558. 

2U  A  grocer  wishes  to  put  840  pounds  of  tea  into  three  kinds 

of  boxes,  containing  respectively  5,  10,  and  15  pounds,  using  the 


PROBLEMS.  61 

same  number  of  boxes  of  each  kind.     How  many  boxes  can  be 
fill?  Ans.  84. 

22.  A  coal  dealer  paid  $965  for  some  coal.  He  sold  160  tons 
for  $5  a  ton,  when  the  remainder  stood  him  in  but  $3  a  ton.  How 
many  tons  did  he  buy?  Ans.  215. 

23.  A  dealer  in  horses  gave  $7560  for  a  certain  number,  and 
sold  a  part  of  them  for  $3825,  at  $85  each,  and  by  so  doing,  lost 
$5  a  head ;  for  how  much  a  head  must  he  sell  the  remainder,  to 
gain  $945  on  the  whole  ?  Ans.  $120. 

24.  Bought  a  Western  farm  for  $22,360,  and  after  expending 
$1742  in  improvements  upon  it,  I  sold  one  half  of  it  for  $15480, 
at  $18  per  acre.  How  many  acres  of  land  did  I  purchase,  and  at 
what  price  per  acre  ? 

PROBLEMS  IN  SIMPLE  INTEGRAL  NUMBERS. 

134«  The  four  operations  that  have  now  been  considered,  viz.. 
Addition,  Subtraction,  Multiplication,  and  Division,  are  all  the 
operations  that  can  be  performed  upon  numbers,  and  hence  they 
are  called  the  Fundamental  Eules. 

l^O.  In  all  cases,  the  numbers  operated  upon  and  the  results 
obtained,  sustain  to  each  other  the  relation  of  a  whole  to  its  parts. 
Thus, 

I.  In  Addition^  the  numbers  added  are  the  parts,  and  the  sum 
or  amount  is  the  whole. 
II.  In   Sifhtraction,  the    subtrahend    and    remainder  are  the 

parts,  and  the  minuend  is  the  whole. 
III.  In  Multiplicationy  the  multiplicand  denotes  the  value  of  one 
part,  the  multiplier  the  number  of  parts,  and  the  pro- 
duct the  total  value  of  the  whole  number  of  parts. 
lY.  In  Division,  the  dividend  denotes  the  total  value  of  the 
whole  number  of  parts,  the  divisor  the  value  of  one 
part,  and  the  quotient  the  number  of  parts ;  or  the 
divisor  the  number  of  parts,  and  the  quotient  the 
value  of  one  part. 
1S6.    Every  example  that  can  possibly  occur  in  Arithmetic, 
and  every  business  computation  requiring  an  arithmetical  opera- 
6 


62  SIMPLE  NUMBERS. 

tion,  can  be  classed  under  one  or  more  of  the  four  Fundamental 
Rules,  as  follows : 

I.  Cases  requiring  Addition. 

There  may  he  given  To  find 

1.  The  parts,  the  whole,  or  the  sum  total. 

2    The  less  of  two  numbers  and  ^ 

^,    .     ,.«.  ^1         1      I  the    g-reater  number  or  the 

their  dmerence,  or  the  sub-   >  P 

^    ,       1       T  .    T  minuend. 

trahend  and  remainder,  J 

II.  Cases  requiring  Subtraction. 
There  may  he  (jiven  To  find 

1.  The  sum  of  two  numbers  and  ) 

one  of  them,  j  *^^  °t^e'^- 

2.  The  greater  and  the  less  of  ^ 

two  numbers,  or  the  minuend  I  the  difference  or  remainder 
and  subtrahend,  J 

III.  Cases  requiring  Multiplication. 
There  may  he  given  To  find 

1.  Two  numbers,  their  product. 

"2.  Any  number  of  factors,  their  continued  product. 

3.  The  divisor  and  quotient,  the  dividend. 

IV.  Cases  requiring  Division. 
There  may  he  given  To  find 

1.  The  dividend  and  divisor,  the  quotient. 

2.  The  dividend  and  quotient,       the  divisor. 

3.  The  product  and  one  of  two  ' 
factors, 

4.  The  continued   product   of  ^ 

several  factors,  and  the  pro-  >  that  one  factor. 
duct  of  all  but  one  factor,     J 

I2T.  Let  the  pupil  be  required  to  illustrate  the  following  pro- 
blems by  orio^inal  examples. 

Problem  1.  Given,  several  numbers,  to  find  their  sum. 

Prob.  2.  Given,  the  sum  of  several  numbers  and  all  of  them 
but  one,  to  find  that  one. 


V  the  other  factor. 


^KsmF 


PROBLEMS.  g3 


Prob.  3.  Given,  the  parts,  to  find  the  whole. 

Prob.  4.  Given,  the  whole  and  all  the  parts  but  one,  to  find 
that  one. 

Prob.  5.  Given,  two  numbers,  to  find  their  difierence. 

Prob.  6.  Given,  the  greater  of  two  numbers  and  their  difierence, 
to  find  the  less  number. 

Prob.  7.  Given,  the  less  of  two  numbers  and  their  difference,  to 
find  the  greater  number. 

Prob.  8.  Given,  the  minuend  and  subtrahend,  to  find  the 
remainder. 

Prob.  9.  Given,  the  minuend  and  remainder,  to  find  the  sub- 
trahend. 

Prob.  10.  Given,  the  subtrahend  and  remainder,  to  find  the 
minuend. 

Prob.  11.  Given,  two  or  more  numbers,  to  find  their  product. 

Prob.  12.  Given,  the  product  and  one  of  two  factors,  to  find  the 
other  factor. 

Prob.  13.  Given,  the  continued  product  of  several  factors  and 
all  the  factors  but  one,  to  find  that  factor. 

Prob.  14.  Given,  the  factors,  to  find  their  product. 

Prob.  15  Given,  the  multiplicand  and  multiplier,  to  find  the 
product. 

Prob.  16.  Given,  the  product  and  multiplicand,  to  find  the 
multiplier. 

Prob.  17.  Given,  the  product  and  multiplier,  to  find  the  mul- 
tiplicand. 

Prob.  18.  Given,  two  numbers,  to  find  their  quotients. 

Prob.  19.  Given,  the  divisor  and  dividend,  to  find  the  quotient. 

Prob.  20.  Given,  the  divisor  and  quotient,  to  find  the  dividend. 

Prob.  21.  Given,  the  dividend  and  quotient,  to  find  the  divisor. 

Prob.  22.  Given,  the  divisor,  quotient,  and  remainder,  to  find 
the  dividend. 

Prob.  23.  Given,  the  dividend,  quotient,  and  remainder,  to  find 
the  divisor. 

Prob.  24.  Given,  the  final  quotient  of  a  continued  division  and 
the  several  divisors,  to  find  the  dividend. 


64  SIMPLE  NUMBEKS. 

Prob  25.  Given,  the  final  quotient  of  a  continued  division,  the 
first  dividend,  a,nd  all  the  divisors  but  one,  to  find  that  divisor. 

Prob.  26.  Given,  the  dividend  and  several  divisors  of  a  con- 
tinued division,  to  find  the  quotient. 

Prob.  27.  Given,  two  or  more  sets  of  numbers,  to  find  the 
difierence  of  their  sums. 

Prob.  28.  Given,  two  or  more  sets  of  factors,  to  find  the  sum  of 
their  products. 

Prob.  29.  Given,  one  or  more  sets  of  factors  and  one  or  more 
numbers,  to  find  the  sum  of  the  products  and  the  given  numbers. 

Prob.  30.  Given,  two  or  more  sets  of  factors,  to  find  the  ditler- 
ence  of  thoir  products. 

Prob.  31.  Given^  one  or  more  sets  of  factors  and  one  or  more 
numbers,  to  find  the  sum  of  the  products  and  the  given  number 
or  numbers. 

Prob.  32.  Given,  two  or  more  sets  of  factors  and  two  or  more 
other  sets  of  factors,  to  find  the  difference  of  the  sums  of  the 
products  of  the  former  and  latter. 

Prob.  33.  Given,  the  sum  and  the  difference  of  two  numbers,  to 
find  the  numbers. 

Analysis.  If  the  difference  of  two  unequal  numbers  be  added  to 
the  less  number,  the  sum  wdll  be  equal  to  the  greater ;  and  if  this 
sum  be  added  to  the  greater  number,  the  result  will  be  twice  the 
greater  number.  But  this  result  is  the  sum  of  the  two  numbers  plus 
their  difference. 

Again,  if  the  difference  of  two  numbers  be  subtracted  from  the 
greater  number,  the  remainder  will  be  equal  to  the  less  number ;  and 
if  this  remainder  be  added  to  the  less  number,  the  result  will  be  twice 
the  less  number.  But  this  result  is  the  sum  of  the  two  numbers 
minus  their  difference.     Hence, 

I.  The  sum  of  two  numbers  plus  their  difference  is  equal  to 
twice  the  greater  number. 

II.  The  sum  of  two  numbers  minus  their  difference  is  equal  to 
twice  the  less  number. 


EXACT  DIVISORS.  65 


PROPERTIES  OF  NUMBERS. 

EXACT  DIVISORS. 

1SS8.  An  Exact  Divisor  of  a  number  is  one  that  gives  an 
integral  number  for  a  quotient.  And  since  division  is  the  reverse 
of  multiplication,  it  follows  that  all  the  exact  divisors  of  a  number 
are  factors  of  that  number,  and  that  all  its  factors  are  exact 
divisors. 

Notes. — 1.  Every  number  is  divisible  by  itself  and  unity ;  but  the  number 
itself  and  unity  are  not  generally  considered  as  factors,  or  exact  divisors  of  the 
number.  '^ 

2.  An  exact  divisor  of  a  number  is  sometimes  called  the  measure  of  the 
number. 

ISO ,  An  Even  Number  is  a  number  of  which  2  is  an  exact 
divisor;  as  2,  4,  6,  or  8. 

130*  An  Odd  Number  is  a  number  of  which  2  is  not  an  exact 
divisor;  as  1,  3,  5,  7,  or  9. 

131.  A  Perfect  Number  is  one  that  is  equal  to  the  sum  of 
all  its  factors  plus  1;  as  6  =  3  +  2  +  1,  or  28  =  14 +  7  +  4  + 

2  +  1. 

Note  —The  only  perfect  numbers  known  are  6,  28,  496,8128,  33550336, 
8589869056,  137438691328,  2305843008139952128,  2417851639228158837784576, 
990352031428297183044881 6128. 

133.  An  Imperfect  Number  is  one  that  is  not  equal  to  the 
sum  of  yi  its  factors  plus  1 ,  as  12,  which  is  not  equal  to  6  +  4 
+  3  +  2  +  1. 

133.  An  Abundant  Number  is  one  which  is  less  than  the 
sum  of  all  its  factors  phis  1 ;  as  18,  which  is  less  than  9  +  6  + 

3  +  2  +  1. 

134.  A  Defective  Number  is  one  which  is  greater  than  the 
sum  of  all  its  factors  plus  1 ;  as  27^  which  is  greater  than  9  +  3  +  1. 

13o.    To  show  che  nature  of  exact  division,  and  furnish  tests 

of  divisibility,  observe  that  if  we  begin  with  any  number,  as  4, 

and  take  once  4,  two  times  4,  three  times  4,  four  times  4,  and  so 

on  indefinitely,  forming  the  series  4,  8,  12,  16,  etc.,  we  shall  have 

6*  B 


66  PROPERTIES  OF  NUMBERS. 

all  the  numbers  that  are  divisible  by  4 ;  and  from  the  manner  of 
forming  this  series,  it  is  evident, 

1st.  That  the  product  of  any  one  number  of  the  series  by  any 
integral  number  whatever,  will  contain  4  an  exact  number  of 
times ; 

2d.  The  sum  of  any  two  numbers  of  the  series  will  contain  4 
an  exact  number  of  times ;  and 

8d.  The  difference  of  any  two  will  contain  4  an  exact  number 
of  times.     Hence, 

I  Any  number  which  will  exactly  divide  one  of  two  numbers 
will  divide  their  product. 

II.  Any  number  which  will  exactly  divide  each  of  two  numbers 
will  divide  their  sum. 

III.  Any  number  which  will  exactly  divide  each  of  two  num- 
bers will  divide  their  difference. 

136.    From  these  principles  we  derive  the  following  properties : 

I.  Any  number  terminating  with  0,  00,  000,  etc.,  is  divisible 
by  10,  100,  1000,  etc.,  or  by  any  factor  of  10,  100,  or  1000. 

For  by  cutting  off  the  cipher  or  ciphers,  the  number  will  be  divided 
by  10,  100.  or  1000,  etc.,  without  a  remainder,  (122)  ;  and  a  number 
of  which  10,  100,  or  1000,  etc.,  is  a  factor,  will  contain  any  factor  of 
10,  100,  or  1000,  etc.,  (I). 

II.  A  number  is  divisible  by  2  if  its  right  hand  figure  is  even 
or  divisible  by  2. 

For,  the  part  at  the  left  of  the  units'  placed  taken  alone,  with  its 
local  value,  is  a  number  which  terminates  with  a  cipher,  and  is  divi- 
sible by  2,  because  2  is  a  factor  of  10,  (I)  ;  and  if  both  parts,  taken 
separately,  with  their  local  values,  are  divisible  by  2,  their  sum,  which 
is  the  entire  number,  is  divisible  by  2,  (135,  II). 

Note. — Hence,  all  numbers  terminating  with  0,  2,  4,  6,  or  8,  are  even,  and  all 
numbers  terminating  with  1,  3,  5,  7,  or  9,  are  odd. 

III.  A  number  is  divisible  by  4  if  the  number  expressed  by  its 
two  right  hand  figures  is  divisible  by  4. 

For,  the  part  at  the  left  of  the  tens^  place,  taken  alone,  with  its 
local  value,  is  a  number  which  terminates  with  two  ciphers,  and  is 
divisible  by  4,  because  4  is  a  factor  of  100,  (I)  ;  and  if  both  parts, 


EXACT  DIVISORS.  gf 

taken  separately,  with  their  local  values,  are  divisible  by  4,  their  sum 
which  is  the  entire  number,  is  divisible  by  4,  (135,  II) 

TV.  A  number  is  divisible  by  8  if  the  number  expressed  by  its 
three  right  hand  figures  is  divisible  by  8. 

For,  the  part  at  the  left  of  the  hundreds'  place,  taken  alone,  with 
its  local  value,  is  a  number  which  terminates  with  three  ciphers,  and 
IS  divisible  by  8,  because  8  is  a  factor  of  1000,  (I) ;  and  if  both 
parts,  taken  separately,  with  theiv  local  values,  are  divisible  by  8, 
their  sum,  or  the  entire  number,  .s  divisible  by  8,  (135,  II). 

Y.  A  number  is  divisible  by  any  power  of  2,  if  as  many  right 
hand  figures  of  the  number  as  are  equal  to  the  index  of  the  given 
power^  are  divisible  by  the  given  power. 

For,  as  2  is  a  factor  of  10,  any  power  of  2  is  a  factor  of  the  corres- 
ponding power  of  10,  or  of  a  unit  of  an  order  one  higher  than  is 
indicated  by  the  index  of  the  given  power  of  2 ;  and  if  both  parts 
of  a  number,  taken  separately,  with  their  local  values,  are  divisible 
by  a  power  of  2,  their  sum,  or  the  entire  number,  is  divisible  by  the 
same  power  of  2,  (135,  II). 

YI.  A  number  is  divisible  by  5  if  its  right  hand  figure  is  0, 
or  5. 

For,  if  a  number  terminates  with  a  cipher,  it  is  divisible  by  5, 
because  5  is  a  factor  of  10,  (I) ;  and  if  it  terminates  with  5,  both 
parts,  the  units  and  the  figures  at  the  left  of  units,  taken  separately, 
with  their  local  values,  are  divisible  by  5,  and  consequently  their 
sum,  or  the  entire  number,  is  divisible  by  5,  (135,  II). 

YII.  A  number  is  divisible  by  25  if  the  number  expressed  by 
its  two  right  hand  figures  is  divisible  by  25. 

For,  the  part  at  the  left  of  the  tens'  figure,  taken  with  its  local 
value,  is  a  number  terminating  with  two  ciphers,  and  is  divisible  by 
25,  because  25  Is  a  factor  of  100,  (I)  ;  and  if  both  parts,  taken 
separately,  with  their  local  values,  are  divisible  by  25,  their  sum,  or 
the  entire  number,  is  divisible  by  25,  (135,  II). 

YIII.  A  number  is  divisible  by  any  power  of  5,  if  as  many 
right  hand  figures  of  the  number  as  are  equal  to  the  index  of  the 
given  power  are  divisible  by  the  given  power. 

For,  as  5  is  a  factor  of  10,  any  power  of  5  is  a  factor  of  the  corres- 
ponding power  of  10,  or  of  a  unit  of  an  order  one  higher  than  is  indi- 


J58  PROPERTIES  OF  NUMBERS. 

cated  by  the  index  of  the  given  power  of  5 ;  and  if  both  parts  of  a 
number,  taken  separately,  with  their  local  values,  are  divisible  by  a 
power  of  5,  their  sum,  or  the  entire  number,  is  divisible^by  the  same 
power  of  5,  (135,  II). 

IX.  A  number  is  divisible  by  9  if  the  sum  of  its  digits  is  divis- 
ible by  9. 

For,  if  any  number,  as  7245,  be  separated  into  its  parts,  7000  -f 
200  +40+5,  and  each  part  be  divided  by  9,  the  several  remainders 
will  be  the  digits  7,  2,  4,  and  5,  respectively ;  hence,  if  the  sum  of 
these  digits,  or  remainders,  be  9  or; an  exact  number  of  9^s,  the  entire 
number  must  contain  an  exact  number  of  9's,  and  will  therefore  be 
divisible  by  9. 

Note.  —  Whence  it  follows  that  if  a  number  be  divided  by  9,  the  remainder 
will  be  the  same  as  the  excess  of  9's  in  the  sum  of  the  digits  of  the  number. 
Upon  this  property  depends  one  of  the  methods  of  proving  the  operations  in 
the  four  Fundamental  Rules. 

X.  A  number  is  divisible  by  a  composite  number,  when  it  is 
divisible,  successively,  by  all  the  component  factors  of  the  com- 
posite number. 

For,  dividing  any  number  successively  by  several  factors,  is  the 
same  as  dividing  by  the  product  of  these  factors,  (119,  I). 

XI.  An  odd  number  is  not  divisible  by  an  even  number. 

For,  the  product  of  any  even  number  by  any  odd  number  is  even , 
and,  consequently,  any  composite  odd  number  can  contain  only  odd 
factors. 

XII.  An  even  number  that  is  divisible  by  an  odd  number,  is 
also  divisible  by  twice  that  odd  number. 

For,  if  any  even  number  be  divided  by  an  odd  number,  the  quo- 
tient must  be  even,  and  divisible  by  2 ;  hence,  the  given  even  num- 
ber, being  divisible  successively  by  the  odd  number  and  2,  will  be 
divisible  by  their  product,  or  twice  the  odd  number,  (119,  I). 

PRIME  NUMBERS. 

1S7.  A  Prime  Number  is  one  that  can  not  be  resolved  or 
separated  into  two  or  more  integral  factors. 

Note.  —  Every  number  must  be  either  prime  or  composite. 


PRIME  NUMBERS. 


C9 


138*  To  find  all  the  prime  numbers  within  any  given  limit, 
we  observe  that  all  even  numbers  except  2  are  composite ;  hence, 
the  prime  numbers  must  be  sought  among  the  odd  numbers. 

130..  If  the  odd  numbers  be  written  in  their  order,  thus;  1, 
3,  5,  1,  9,  11,  13,  15  17,  etc.,  we  observe, 

1st.  Taking  every  third  number  after  3,  we  have  3  times  3,  5 
times  3,  7  times  3,  and  so  on ;  which  are  the  only  odd  numbers 
divisible  by  3. 

2d.  Taking  every  fifth  number  after  5,  we  have  3  times  6,  5 
times  5,  7  cimes  5,  and  so  on;  which  are  the  only  odd  numbers 
divisible  by  5.  And  the  same  will  be  true  of  every  other  number 
in  the  series.     Hence, 

3d.  If  we  cancel  every  third  number,  counting  from  3,  no 
number  divisible  by  3  will  be  left;  and  since  3  times  5  will  be 
canceled,  5  times  5.  or  25,  will  be  the  least  composite  number  left 
in  the  series.     Hence^ 

4th.  If  we  cancel  every  fifth  number,  counting  from  25,  no 
number  divisible  by  5  will  be  left;  and  since  3  times  7,  and  5 
times  7,  will  be  canceled,  7  times  7,  or  49,  will  be  the  least  com- 
posite number  left  in  the  series.  And  thus  with  all  the  prime 
numbers.     Hence, 

140.  To  find  all  the  prime  numbers  within  any  given  limit, 
we  have  the  following 

Rule.     I.    Write  all  the  odd  riumhers  in  their  natural  order. 

II.  Cancel,  or  cross  out,  3  times  3^  or  9,  and  every  third  number 
after  it;  5  times  5,  or  25,  and  every  fifth  number  after  it;  7  times 
7,  or  49,  and  every  seventh  number  after  it ;  and  so  on,  beginning 
with  the  second  power  of  each  prime  number  in  succession,  till  the 
given  limit  is  reached.  The  numbers  remaining,  together  tvith  the 
number  2,  will  be  the  prime  numbers  required. 

Notes. — 1.  It  is  unnecessary  to  count  for  every  ninth  number  after  9  times  9, 
for  being  divisible  by  3,  they  will  be  found  already  canceled;  the  same  may  be 
said  of  any  other  canceled,  or  composite  number. 

2.  This  method  of  obtaining  a  list  of  the  prime  numbers  was  employed  by 
Eratosthenes  (born  b.  c,  275),  and  is  called  Eratoathenea'  Sieve, 


70  PROPERTIES  OF  NUMBERS. 

TABLE  OF  PRIME  NUMBERS  LESS  THAN  1000. 


1 

59 

139 

233 

337 

439 

557 

653 

769 

883 

2 

61 

149 

239 

347 

443 

563 

659 

773 

887 

3 

67 

151 

241 

349 

449 

569 

661 

787 

907 

6 

71 

157 

251 

353 

457 

571 

673 

797 

911 

7 

73 

163 

257 

359 

461 

577 

677 

S09 

919 

11 

79 

167 

263 

367 

463 

587 

683 

811 

929 

13 

83 

173 

269 

37d 

467 

593 

691 

821 

937 

17 

89 

179 

271 

379 

479 

599 

701 

823 

941 

19 

97 

181 

277 

383 

487 

601 

709 

827 

947 

23 

101 

191 

281 

389 

491 

607 

719 

829 

953 

29 

103 

193 

283 

397 

499 

613 

727 

839 

967 

31 

107 

197 

293 

401 

503 

617 

733 

853 

971 

37 

109 

]99 

307 

409 

509 

619 

739 

857 

977 

41 

113 

211 

311 

419 

521 

631 

743 

859 

983 

43 

127 

223 

313 

421 

523 

641 

751 

863 

991 

47 

131 

227 

317 

431 

541 

643 

757 

877 

997 

53 

137 

229 

331 

433 

547 

647 

761 

881 

FACTORING. 
CASE  I. 


number   into   its 


141,    To   resolve   any    composite 
prime  factors. 

The  Prime  Factors  of  a  number  are   those  prime  numbers 
which  multiplied  together  will  produce  the  given  number. 

14:2,    The  process  of  factoring  numbers  depends  upon  the  fol- 
lowing principles : 

I.  Every  prime  factor  of  a  number  is  an  exact  divisor  of  that 
number. 

II.  The  only  exact  divisors  of  a  number  are  its  prime  factors,  or 
some  combination  of  its  prime  factors. 

1.  What  are  the  prime  factors  of  798  ? 

OPERATION. 


2 
3 

7 
19 


798 
399 
133 
"19 


Analysis.  Since  the  given  number  is  even,  we 
divide  by  2,  and  obtain  an  odd  number,  399,  for  a 
quotient  We  then  divide  by  the  prime  numbers 
3,  7,  and  19,  successively,  and  the  last  quotient  is 
1.  The  divisors,  2,  3,  7,  and  19,  are  the  prime 
factors  required,  (II).     Hence,  the 


FACTORING.  YJ 

Rule.  Divide  the  given  number  hy  any  prime  factor ;  divide 
the  quotient  in  the  same  manner ,  and  so  continue  the  division  until 
the  quotient  is  a  prime  number.  The  several  divisors  and  the  last 
quotient  will  be  the  prime  factors  required. 

Proof.  The  product  of  all  the  prime  factors  wiP  be  the  given 
number. 

EXAMPLES    FOR   PRACTICE. 

1.  What  arc  the  prime  factors  of  2150  'Z 

2.  What  are  the  prime  factors  of  2445? 

3.  What  are  the  prime  factors  of  6300  ? 

4.  What  are  the  prime  factors  of  21504? 

5.  What  are  the  prime  factors  of  2366  ? 

6.  What  are  the  prime  factors  of  1000  ? 

7.  What  are  the  prime  factors  of  390625? 

8.  What  are  the  prime  factors  of  999999  ? 

143.  If  the  prime  factors  of  a  number  are  small;  as  2,  3,  5; 
7,  or  11,  they  may  be  easily  found  by  the  tests  of  divisibility, 
(136),  or  by  trial.  But  numbers  may  be  proposed  requiring  many 
trials  to  find  their  prime  factors.  This  difficulty  is  obviated, 
within  a  certain  limit,  by  the  Factor  Table  given  on  pages  72,  73. 

By  prefixing  each  number  in  bold-face  type  in  the  column 
of  Numbers,  to  the  several  numbers  following  it  in  the  same  divis- 
ion of  the  column,  we  shall  form  all  the  composite  numbers  less 
than  10,000,  and  not  divisible  by  2,  3,  5,  7,  or  11;  the  numbers 
in  the  columns  of  Factors  are  the  least  prime  factors  of  the  num- 
bers thus  formed  respectively.  Thus,  in  one  of  the  columns  of 
Numbers  we  find  39,  in  bold-face  type,  and  below  39,  in  the  same 
column,  is  77,  which  annexed  to  39,  forms  3977,  a  composite  num- 
ber. The  least  prime  factor  of  this  number  is  41,  which  we  find 
at  the  right  of  77,  in  the  column  of  Factors. 

144.  Hence,  for  the  use  of  this  table,  we  have  the  following 
BULE.     I.    Cancel  from  the  given  number  all  factors  less  than 

13,  and  then  find  the  remaining  factors  by  the  table. 

II.  If  any  number  less  than  10,000  is  not  found,  in  the  tabU, 
and  is  not  divisible  by  2,  3,  5,  7,  or  11,  it  is  prime. 


72 


PROPERTIES  OF  NUMBERS. 


FACTOR   TABLE. 


1 1 

i  ^ 

.a   0 

Numbers. 
Factori. 

1  i 

ll 

1  1 

|1 

1 

99  29 

11  17 

43  29 

79  37 

41  17 

83  17 

41  23 

09  31 

17  53 

77  31 

69  13 

9 

17  13 

57  19 

83  13 

51  13 

97  43 

59  17 

13  19 

27  29 

83  71 

a 

01  17 

57  31 

61  37 

89  19 

77  13 

34: 

69  53 

21  29 

47  47 

91  29 

21  13 

23  13 

69  13 

63  13 

91  47 

83  19 

01  19 

87  13 

31  61 

67  67 

5a 

47  13 

43  23 

15 

ao 

as 

87  29 

03  41 

93  17 

4343 

69  19 

07  41 

89  17 

49  13 

01  19 

21  43 

01  41 

93  41 

19  13 

39 

51  19 

71  13 

13  13 

99  13 

61  31 

13  17 

33  19 

07  23 

30 

27  23 

01  47 

69  17 

77  17 

19  17 

3 

89  23 

17  37 

41  13 

09  13 

07  31 

31  47 

37  31 

79  29 

48 

21  23 

23  17 

10 

37  29 

47  23 

33  17 

13  23 

39  19 

53  59 

81  13 

11  17 

39  13 

61  19 

03  17 

41  23 

59  29 

37  43 

29  13 

73  23 

59  37 

87  41 

19  61 

49  29 

77  13 

07  19 

77  19 

71  19 

61  13 

43  17 

81  59 

61  17 

93  23 

41  47 

51  59 

91  17 

27  13 

91  37 

77  31 

67  17 

53  43 

97  13 

73  29 

99  53 

43  29 

63  19 

4 

37  17 

16 

%\ 

73  31 

71  37 

35 

77  41 

4:4: 

47  37 

67  23 

03  13 

73  29 

33  23 

17  29 

81  29 

77  17 

03  31 

79  23 

27  19 

49  13 

87  17 

37  19 

79  13 

43  31 

19  13 

87  13 

97  19 

23  13 

91  13 

2943 

53  23 

93  67 

81  13 

81  23 

49  17 

47  19 

99  23 

31 

51  53 

40 

39  23 

5943 

53 

93  17 

11 

51  13 

59  17 

ae 

03  29 

69  43 

09  19 

53  61 

67  31 

11  47 

5 

21  19 

79  23 

71  13 

03  19 

07  13 

87  17 

31  29 

69  41 

83  19 

17  13 

27  17 

39  17 

81  41 

7341 

2343 

27  53 

89  37 

33  37 

71  17 

91  67 

21  17 

29  23 

47  31 

91  19 

83  37 

27  37 

31  31 

99  59 

43  13 

89  67 

97  59 

2973 

33  13 

57  13 

IT 

97  13 

41  19 

33  13 

36 

61  31 

45 

49 

39  19 

51  19 

59  19 

03  13 

aa 

69  17 

39  43 

01  13 

63  17 

11  13 

01  13 

53  53 

59  13 

89  29 

11  29 

01  31 

ai 

4947 

11  23 

69  13 

31  23 

13  17 

59  23 

89  19 

la 

17  17 

09  47 

01  37 

51  23 

29  19 

87  61 

37  13 

27  13 

63  31 

6 

07  17 

39  37 

27  17 

4313 

61  29 

4941 

97  17 

41  19 

79  13 

71  41 

11  13 

19  23 

51  17 

31  23 

47  41 

73  19 

53  13 

4:1 

53  29 

81  17 

77  19 

29  17 

41  17 

63  41 

49  13 

59  31 

93  31 

67  19 

17  23 

5947 

97  19 

89  17 

67  13 

47  29 

69  29 

57  37 

71  17 

97  23 

79  13 

21  13 

73  17 

50 

54 

89  13 

61  13 

81  i; 

63  31 

7347 

3a 

83  29 

41  41 

77  23 

17  29 

29  61 

97  17 

71  31 

18 

7943 

as 

11  13 

37 

63  23 

79  19 

2947 

47  13 

7 

73  19 

07  13 

91  29 

09  53 

33  53 

1347 

71  43 

89  13 

41  71 

59  53 

03  19 

13 

17  23 

33 

13  29 

3941 

21  61 

81  37 

46 

53  31 

6143^ 

13  23 

13  13 

19  17 

23  23 

31  19 

47  17 

87  37 

83  47 

01  43 

57  13 

73  13 

31  17 

33  31 

29  31 

27  13 

39  17 

63  13 

43  19 

87  53 

07  17 

63  61 

91  17 

67  13 

39  13 

43  19 

29  17 

67  47 

77  29 

49  23 

89  59 

19  31 

69  37 

97  23 

79  19 

43  17 

4943 

53  13 

69  19 

81  17 

57  13 

99  13 

33  41 

83  13 

55 

93  13 

49  19 

53  17 

63  17 

73  13 

87  19 

63  53 

4:a 

61  59 

51 

13  37 

99  17 

57  23 

91  31 

69  23 

81  43 

93  37 

81  19 

2341 

67  13 

11  19 

39  29 

8 

63  29 

19 

%\ 

99  13 

33 

91  17 

37  19 

81  31 

2347 

43  23 

17  19 

69  37 

09  23 

07  29 

a9 

17  31 

99  29 

47  31 

87  43 

29  23 

49  31 

41  29 

87  19 

19  19 

13  19 

11  41 

37  47 

38  • 

67  17 

93  13 

41  53 

61  67 

61  23 

91  13 

21  17 

19  41 

21 23 

41  13 

09  13 

4:3 

99  37 

43  37 

67  19 

71  ij 

14: 

27  41 

49  31 

23  37 

49  17 

11  37 

03  13 

47 

49  19 

87  37 

93  19 

03  23 

37  13 

61  23 

29  29 

79  31 

27  43 

07  59 

09  17 

61  13 

»7  29 

FACTORING. 


78 


FACTOR   TABLE  — Continued. 


1  s 

a   y 

ii 

»   fx 

2  „• 

1  (2 

2   „• 

1  1 

1      3    c< 

1  1 

li 

li 

;2;  C 

55  ;S 

ll 

1  d 

56 

60 

3947 

51  13 

61  53 

57  13 

23  71 

1347 

51  53 

53  19 

59  13 

03  13 

01  17 

43  17 

59  19 

67  13 

61  47 

27  23 

17  19 

57  17 

59  47 

71  19 

OD  71 

19  13 

63  23 

77  13 

77  19 

63  79 

33  29 

41  23 

73  19 

63  59 

73  17 

11  31 

23  19 

67  29 

87  71 

79  29 

97  43 

47'  13 

53  79 

79  13 

69  13 

83  23 

17  41 

31  37 

87  13 

89  83 

89  37 

77 

51  83 

71  43 

81  83 

71  73 

97 

27  17 

49  23 

93  43 

93  61 

91  23 

09  13 

77  41 

73  37 

91  17 

87  37 

01  89 

29  13 

59  7J 

97  73 

69 

73 

29  59 

83  59 

79  61 

89 

99  17 

03  31 

3343 

71  13 

99  67 

01  67 

03  67 

39  71 

81 

83  17 

03  29 

93 

07  17 

71  53 

77  59 

65 

13  31 

13  71 

47  61 

19  23 

80  13 

09  59 

01  71 

27  71 

81  13 

61 

09  23 

29  13 

19  13 

51  23 

31  47 

97  29 

17  37 

07  41 

31  37 

99  41 

03  17 

11  17 

31  29 

27  17 

69  17 

37  79 

85 

27  79 

13  67 

61  43 

5T 

07  31 

27  61 

43  53 

39  41 

71  19 

43  17 

07  47 

47  23 

22  19 

63  13 

07  13 

09  41 

3347 

53  17 

61  17 

81  31 

49  29 

09  67 

57  13 

47  13 

73  29 

13  29 

19  29 

39  13 

73  19 

63  37 

83  43 

58  31 

81  19 

59  17 

53  ^47 

97  97 

23  59 

37  17 

41  31 

89  29 

67  53 

87  13 

59  41 

49  83 

77  47 

07  17 

99  41 

29  17 

57  47 

57  79 

TO 

73  73 

78 

77  13 

51  17 

83  13 

79  83 

98 

59  13 

61  61 

83  29 

03  47 

79  47 

01  29 

89  19 

57  43 

89  89 

89  41 

09  17 

67  7J 

69  31 

93  19 

0943 

87  83 

07  37 

83 

67  13 

93  17 

94: 

27  31 

71  29 

79  37 

66 

31  79 

91  19 

11  73 

01  59 

79  23 

90 

07  23 

41  13 

73  23 

87  23 

13  17 

33  13 

97  13 

13  13 

03  13 

87  ii 

17  71 

09  97 

4743 

77  5J 

91  41 

17  13 

37  31 

74: 

31  41 

07  29 

93  13 

19  29 

51  13 

53  59 

58 

63 

23  37 

61  23 

09  31 

37  17 

1343 

86 

47  83 

69  17 

69  71 

09  37 

27  13 

31  19 

67  37 

21  41 

49  47 

27  19 

11  79 

61  13 

81  19 

81  41 

33  19 

33  23 

41  29 

81  73 

23  13 

59  29 

49  73 

21  37 

71  47 

87  53 

93  13 

37  13 

39  17 

47  17 

87  19 

29  17 

71  17 

51  37 

33  89 

7343 

95 

99  19 

91  4J 

41  79 

49  61 

93  41 

39  43 

91  13 

57  23 

39  53 

77  29 

03  13 

99 

93  71 

53  13 

67  59 

97  47 

53  29 

97  53 

79  17 

51  41 

83  31 

09  37 

13  23 

99  17 

83  61 

83  41 

99  31 

63  17 

79 

99  43 

53  17 

89  61 

17  31 

1747 

59 

80  19 

97  37 

71 

71  31 

13  41 

83 

71  13 

91 

23  89 

37  19 

09  19 

63 

6r 

11  13 

93  59 

21  89 

03  19 

83  19 

01  19 

29  13 

43  61 

11  23 

13  59 

07  19 

23  17 

75 

8J  17 

21  53 

87 

13  13 

53  41 

53  37 

17  61 

19  71 

31  53 

41  37 

01  13 

43  13 

33  13 

11  31 

31  23 

57  19 

59  23 

21  31 

31  13 

39  23  ■ 

53  23 

19  73 

57  73 

39  31 

17  23 

39  13 

63  73 

71  13 

33  17 

41  17 

49  17 

57  17 

31  17 

61  19 

41  19 

49  13 

4341 

71  17 

79  17 

41  13 

71  23 

51  43 

63  13 

43  19 

67  31 

47  17 

59  19 

67  89 

77  61 

83  67 

47  19 

S3  13 

57  29 

69  67 

71  67 

69  13 

57  61 

73  31 

69  53 

S9  43 

91  97 

59  59 

64: 

67  67 

71  71 

97  71 

79  79 

59  13 

77  67 

79  67 

93  53 

97  13 

63  67 

01  37 

73  13 

81  43 

76 

81  23 

81  17 

91  59 

93  29 

99  29 

69  47 

03  19 

99  13 

99   23 

13  23 

91  61 

83  83 

97  19 

97  17 

96 

7743 

07  4? 

68 

73 

19  19 

99  19 

99  37 

88 

93 

07  13 

83  31 

09  13 

17  17 

01  19 

27  29 

80 

84: 

01  13 

11  61 

17  59 

89  53 

31  59 

21  19 

23  31 

31  13 

03  53 

01  31 

09  23 

17  13 

37  23 

93  13 

37  41 

47  41 

41  13 

33  17 

21  13 

11  41 

43  37 

23  23 

41  31 

3 

188139 

7 

62713 

17 

8959 

17 

527 

31 

74  PROPERTIES  OF  NUMBERS. 

1.  Eesolve  1961  into  its  prime  factors. 

OPERATION.  Analysis.     Cutting  off  the   two 

1961  -—  37  =  53  right   hand    figures   of   the    given 

1961  =  37  X  53.  Ans.        number,  and  referring  to  the  table, 

column  No.,  we  find  the  other  part, 

19,  in  bold-face  type ;  and  under  it,  in  the  same  division  of  the  column, 

we  find  61,  the  figures  cut  off;  at  the  right  of  61,  in  column  Fac,  we 

find  37,  the  least  prime  factor  of  the  given  number.     Dividing  by  37, 

we  obtain  53,  the  other  factor. 

2.  Resolve  188139  into  its  prime  factors. 

OPERATION.  Analysis.     We   find    by   trial 

that  the  given  number  is  divisible 
by  3  and  7  ;  dividing  by  these  fac- 
tors, we  have  for  a  quotient  8959. 
By  referring  to  the  factor  table, 
we  find  the  least  prime  factor  of 
this  number  to  be  17 ;  dividing  by 
17,  we  have  527  for  a  quotient, 
3x7x17x17x31^  Ans.      Preferring  again  to  the  table,  we 

find  17  to  be  the  least  factor  of 
527,  and  the  other  factor,  31,  is  prime. 

EXAMPLES    FOR   PRACTICE. 

1.  Resolve  18902  into  its  prime  factors.       Ans.  2,  13,  727. 

2.  Resolve  352002  into  its  prime  factors. 

3.  Resolve  6851  into  its  prime  factors. 

4.  Resolve  9367  into  its  prime  factors. 

5.  Resolve  203566  into  its  prime  factors. 

6.  Resolve  59843  into  its  prime  factors. 

7.  Resolve  9991  into  its  prime  factors. 

8.  Resolve  123015  into  its  prime  factors. 

9.  Resolve  893235  into  its  prime  factors. 

10.  Resolve  390976  into  its  prime  factors. 

11.  Resolve  225071  into  its  prime  factors. 

12.  Resolve  81770  into  its  prime  factors. 

13.  Resolve  6409  into  its  prime  factors. 

14.  Resolve  178296  into  its  prime  factors. 

15.  Resolve  714210  into  its  prime  factors. 


FACTORING. 


75 


CASE   II. 

145.  To  find  all  the  exact  divisors  of  a  number. 

It  is  evident  that  all  the  prime  factors  of  a  number,  together 
with  all  the  possible  combinations  of  those  prime  factors,  will  con- 
stitute all  the  exact  divisors  of  that  number,  (142,  II). 

1 .  What  are  all  the  exact  divisors  of  360  ? 

OPERATIOX. 

360  =  1x2x2x2x3x3x5. 


Ans.< 


1 , 

2  , 

4  ,       8 

Combinations  of  1  and  2. 

3  , 
9  , 

6  , 

18  , 

12  ,     24 1 

36  ,     72  1 

(( 

"  1  and  2  and  3. 

5  , 

10  , 

20  ,     40 

(( 

"  land  2  and  5. 

15  , 

45  , 

30  , 
90  , 

60  ,  120  1 
180  ,  360  f 

ti 

«  land  2  and  3 

id  5. 


Analysis.  Bj  Case  I  we  find  the  prime  factors  of  360  to  be  1,  2, 
2,  2,  3,  3,  and  5.  As  2  occurs  three  times  as  a  factor,  the  different 
combinations  of  1  and  2  by  which  360  is  divisible  will  be  1,  1x2  =  2, 
1  X  2  X  2  =  4,  and  1  X  2  X  2  X  2  =  8;  these  we  write  in  the  first  line. 
Multiplying  the  first  line  by  3  and  writmg  the  products  in  the  second 
line,  and  the  second  line  by  3,  writing  the  products  in  the  third  line, 
we  have  in  the  first,  second  and  third  lines  all  the  different  combina- 
tions of  1,  2,  and  3,  by  which  360  is  divisible.  Multiplying  the  first, 
second  and  third  lines  by  5,  and  writing  the  products  in  the  fourth, 
fifth  and  sixth  lines,  respectively,  we  have  in  the  six  lines  together, 
every  combination  of  the  prime  factors  by  which  the  given  number, 
360,  is  divisible. 

Hence  the  following 

Rule.     I.  Resolve  the  given  mimher  into  its  prime  /actors, 

II  Form  a  series  having  \  for  the  firsts  term,  that  prime  factor 
which  occurs  the  greatest  number  of  times  in  the  given  numher  for 
the  second  term,  the  square  of  this  factor  for  the  third  term,  and  so 
on,  till  a  term  is  reached  containing  this  factor  as  many  times  as  it 
occurs  in  the  given  number. 

III.  Multiply  the  numbers  in  this  line  by  another  factor,  and 
these  residts  by  the  same  factor,  and  so  on,  as  many  times  as  this 
factor  occurs  in  the  given  number. 


76  PROPERTIES  OF  NUMBERS. 

TV.  Multiply  all  the  comhinations  now  obtained  hy  another 
factor  in  continued  multiplication ^  and  thus  proceed  till  all  the  dif- 
ferent factors  have  been  used.  All  the  combinations  obtained  will 
be  the  exact  divisors  scni<jht. 

EXAMPLES   FOR   PRACTICE. 

1.  What  are  all  the  exact  divisors  of  120  ? 

Ans.  1,  2, 3, 4,  5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. 

2.  Find  all  the  exact  divisors  of  84. 

Ans.  1,  2,  3,  4,  6,  7,  12,  14,  21,  28,  42,  84. 

3.  Find  all  the  exact  divisors  of  100. 

Ans.  1,  2,  4,  5,  10,  20,  25,  50,  100. 

4.  Find  all  the  exact  divisors  of  420. 


^^^  ( 1,  2,  3,  4,  5,  6,  7,  10,  12,  14,  15,  20,  21,  28, 
*  1  30,  35,  42,  60,  70,  84,  105,  140,  210,  420. 
d  all  the  exact  divisors  of  1050. 

r  1,  2,  3,  5,  6,  7,  10,  14,  15,  21,  25,  30,  35,  42,  50,  70, 
1  75,  105,  150,  175,  210,  350,  525,  1050. 


GREATEST  COMMON  DIVISOR. 

94G*  A  Common  Divisor  of  two  or  more  numbers  is  a  number 
that  will  exactly  divide  each  of  them. 

147.  The  Greatest  Common  Divisor  of  two  or  more  numbers 
is  the  greatest  number  that  will  exactly  divide  each  of  them. 

148.  Numbers  Prime  to  each  other  are  such  as  have  no  com- 
mon divisor. 

Note. — A  common  divisor  is  sometimes  called  a  Common  Measure ;  and  the 
greatest  common  divisor,  the  Greatest  Common  Measure. 

CASE   I. 

149.  When  the  numbers  can  be  readily  factored. 

It  is  evident  that  if  several  numbers  have  a  common  divisor, 
they  may  all  be  divided  by  any  component  factor  of  this  divisor, 
and  the  resulting  quotients  by  another  component  factor,  and  so 
on,  till  all  the  component  factors  have  been  used. 


OPERATION. 
28   .   .   140   . 

420 

4  .  .    20  . 

60 

1  .  .       5  . 

15 

GREATEST  COMMON  DIVISOR.  77 

I.  What  is  the  greatest  common  divisor  of  28,  140,  and  420  ? 

Analysis.  We  readily  see  that  7 
will  exactly  divide  each  of  the  given 
numbers ;  and  then,  4  will  exactly 
divide  each  of  the  resulting  quotients. 
Hence,  each  of  the  given  numbers 
4x7  =  28,  Ans.  can  be  exactly  divided  by  7  times  4; 
and  these  numbers  must  be  compo- 
nent factors  of  the  greatest  common  divisor.  Now,  if  there  were  any 
other  component  factor  of  th"e  greatest  common  divisor,  the  quotients, 
1,  5  and  15,  would  be  divisible  by  it.  But  these  quotients  are  prime 
io  each  other ;  therefore,  7  and  4  are  all  the  component  factors  of  the 
greatest  common  divisor  sought. 

From  this  analysis  we  derive  the  following 

Rule.     I.    Write  the  numbers  in  a  liney  with  a  vertical  line  at 
the  left^  and  divide  hy  ariy  factor  common  to  all  the  numbers. 

II.  Divide  the  quotients  in  like  manner,  and  continue  the  divi- 
sion till  a  set  of  quotients  is  obtained  that  are  prime  to  each  other. 

II [.  Multiply  all  the  divisors  together ,  and  the  product  will  be 
the  greatest  common  divisor  sought. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  40,  75,  and  100? 

Ans.  5. 

2.  What  is  the  greatest  common  divisor  of  18,  80,  36,  42, 
and  54  ? 

3.  What  is  the  greatest  common  divisor  of  42,  63,  126,  and 
189?  Ans.  21. 

4.  What  is  the  greatest  common  divisor  of  135,  225,  270,  and 
315?  Ans.  45. 

5.  What  is  the  greatest  common  divisor  of  84,  126,  210,  252, 
294,  and  462  ? 

6.  What  is  the  greatest  common  divisor  of  216,  360,  432,  648, 
and  936?  Ans.  72. 

7.  What  is  the  greatest  common  divisor  of  102,  153,  and  255  ? 

Ans.  51. 
7* 


78  PROPERTIES  OF  NUMBERS. 

8.  What  is  the  greatest  common  divisor  of  756,  and  1575  ? 

9.  What  is  the  greatest  common  divisor  of  182, 364,  and  455? 

10.  What  is  thQ  greatest  common  divisor  of  2520,  and  3240  ? 

Alls.  360. 

11.  What  is  the  greatest  common  divisor  of  1428,  and  1092  ? 

12.  What  is  the  greatest  common  divisor  of  1008,  and  1036? 

Ans,  28. 

CASE    IT. 

1^0.  When  the  numbers  cannot  be  readily  factored. 
The   analysis   of  the   method   in   this   case   depends  upon   the 
following  properties  of  divisors. 

I.  An  exact  divisor  divides  any  number  of  times  its  dividend. 

II.  A  common  divisor  of  two  numbers  is  an  exact  divisor  of 
their  sum. 

III.  A  common  divisor  of  two  numbers  is  an  exact  divisor  of 
their  clifference. 

Note. — The  last  two  properties  are  essentially  the  same  as  102,  II,  III. 

1.  What  is  the  greatest  common  divisor  of  527,  and  1207? 

Analysis.  We  wall  first  describe  the  pro- 
cess, and  then  examine  the  reasons  for  the 
several  steps  in  the  operation.  Drawing  two 
vertical  lines,  we  place  the  greater  number 
on  the  right,  and  the  less  number  on  the  left, 
one  line  lower  down.  We  then  divide  1207, 
the  greater  number,  by  527,  the  less,  and 
w  rite  the  quotient,  2,  between  the  verticals, 
the  product,  1054,  opposite  the  less  number  and  under  the  greater, 
and  the  remainder,  153,  below.  We  next  divide  527  by  this  re- 
mainder, writing  the  quotient,  3,  between  the  verticals,  the  product, 
459,  on  the  left,  and  the*  remainder,  68,  below.  We  again  divide  the 
last  divisor,  153,  by  68,  and  obtain  2  for  a  quotient,  136  for  a  pro- 
duct, and  17  for  a  remainder,  all  of  which  we  write  in  the  same  order 
as  in  the  former  steps.  Finally,  dividing  the  last  divisor,  68,  by  the 
last  remainder,  17,  we  have  no  remainder,  and  17,  the  last  divisor,  is 
the  greatest  common  divisor  of  the  given  numbers. 

Now,  observing  that  the  dividend  is  always  the  s\im  of  the  product 
ftnd  remainder^  and  that  the  remainder  is  always  the  difference  of  the 


0 

PE  RATION. 

1207 

527 

2 

1054 

459 

3 
2 

153 

68 

136 

68 

4 

17 

GREATEST  COMMON  DIVISOR. 


79 


OPERATIOX. 


dividend  and  product,  we  first  trace  the  work  in  the  reverse  order,  as 

indicated  by  the  arrow  line  in  the  diagram  below. 

17  divides  68,  as  proved  by  the 
last  division ;  it  will  also  divide 
2  times  68,  or  136,  (I).  Now,  as 
17  divides  both  itself  and  136,  it 
will  divide  153,  their  sum,  (II). 
It  will  also  divide  3  times  153,  or 
459,  (I)  ;  and  since  it  is  a  com- 
mon divisor  of  459  and  68,  it 
must  divide  their  sum,  527,  which 
is  one  of  the  given  numbers.  It 
will  also  divide  2  times  527,  or 


527 
459 

68 

68 
1054,  (I 


2 

A 

3 

2 

4 

— m^ 

1207 


1054 


153 


136 


17 


and  since  it  is  a  common  divisor  of  1054  and  153,  it  must 
divide  their  sum,  1207,  the  greater  number,  (II).  Hence,  17  is  ^com- 
mon divisor  of  the  given  numbers. 

Again,  tracing  the  work  in  the  direct  order,  as  indicated  in  the  fol- 
lowing diagram,  we  know  that 


527 


459 


68 


1207 


the  ^rea^e^/ common  divisor,  what- 
ever it  he,  must  divide  2  times 
527,  or  1054,  (I).  And  since  it 
will  divide  both  1054  and  1207, 
it  must  divide  their  difference, 
153,  (III).  It  will  also  divide  3 
times  153,  or  459,  (I)  ;  and  as  it 
will  divide  both  459  and  527,  it 
must  divide  their  difference,  68, 
(III).  It  will  also  divide  2  times 
68,  or  136,  (I)  ;  and  as  it  will 
it  must  divide  their  difference,  17,  (III)  ; 


1054 


153 


136 


Y       17 


divide  both  136  and  153^ 

hence,  it  cannot  he  greater  tJian  17. 

Thus  we  have  shown, 

1st.  Tiiat  17  is  a  commmi  divisor  of  the  given  numbers. 
2d.  That  their  greatest  common  divisor,  whatever  it  be,  cannot 
be  greater  than  17.     Hence  it  must  be  17. 

From  this  example  and  analysis,  we  derive  the  following 

Rule.     I.  Draw  two  verticals,  and  write  the  two  riumhers,  ons 
on  each  side,  the  greater  number  one  line  above  the  less. 


80 


PROPERTIES  OF  NUMBERS. 


II.  Divide  the  greater  numher  hy  the  less,  writing  the  quotient 
between  the  verticals^  the  product  wider  the  dividend^  and  the  re- 
mainder  helow. 

III.  Divide  the  less  numher  hy  the  remainder,  the  last  divisor 
hy  the  last  remaindery  and  so  on,  till  nothing  remains.  The  last 
divisor  will  he  the  greatest  common  divisor  sought. 

lY.  If  more  than  two  numhers  he  given,  first  find  the  greatest 
common  divisor  of  two  of  them,  and  then  of  this  divisor  and  one 
of  the  remaining  numhers,  and  so  on  to  the  last ;  the  last  common 
divisor  found  will  he  the  greatest  common  divisor  required. 

Notes. — 1.  "When  more  than  two  numbers  are  giren,  it  is  better  to  begin  with 
the  least  two. 

2.  If  at  any  point  in  the  operation  a  prime  number  occur  as  a  remainder,  it 
must  be  a  common  divisor,  or  the  given  numbers  have  no  common  divisor. 


EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  18607  and  417979? 

OPERATION. 

417979 

18607 


17250 


1357 

966 


391 

868 


Ans. 


23 


2 

37214 

45839 

2 

37214 

2 

8625 

6 

8142 

2 

483 

1 

391 

4 

92 

4 

92 

2.  What  is  the  greatest  common  divisor  of  10661  and  12303? 


OPERATIOX. 

12303 


10661 
9852 


Prime       809 


10661 


1642 


Ans.  1. 


GREATEST  COMMON  DIVISOR. 


81 


3.  What  is  the  greatest  common  divisor  of  336  and  812  ? 

Ans.   28. 

4.  What  is  the  greatest  common  divisor  of  407  and  1067  ? 

5.  What  is  the  greatest  common  divisor  of  8J5  and  1:>72  ? 

.        6.  What  is  the  greatest  common  divisor  of  2041  and  8476  ? 
^  Ans.   13. 

7.  What  is  the  greatest  common  divisor  of  3281  and  10778? 

8.  Find  the  greatest  common  divisor  of  22579,  and  116939. 

9.  What  is  the  greatest  common  divisor  of  49373  and  147731  ? 

Ans.  97. 

10.  What  is  the  greatest  common  divisor  of  1005973  and 
4616175? 

11.  Find  the  greatest  common  divisor  of  292^  1022,  and  1095. 

Ans.  73. 

12.  What  is  the  greatest  common  divisor  of  4718,  6951,  and 
8876?  Ans.  7. 

13.  Find  the  greatest  common  divisor  of  141,  799,  and  940. 

14.  What  is  the  greatest  common  divisor  of  484391  and  684877  ? 

Ans.   701. 

15.  A  farmer  wishes  to  put  364  bushels  of  corn  and  455  bushels 
of  oats  into  the  least  number  of  bins  possible,  that  shall  contain 
the  same  number  of  bushels  without  mixing  the  two  kinds  of 
grain ;  what  number  of  bushels  must  each  bin  hold  ? 

A71S.   91. 

16.  A  gentleman  having  a  triangular  piece  of  land,  the  sides  of 
which  are  165  feet,  231  feet,  and  385  feet,  wishes  to  inclose  it 
with  a  fence  having  pannels  of  the  greatest  possible  uniform 
length;  what  will  be  the  length  of  each  pannel? 

17.  B  has  $620,  C  $1116,  and  D  $1488,  with  which  they  agree 
to  purchase  horses,  at  the  highest  price  per  head  that  will  allow 
each  man  to  invest  all  his  money;  how  many  horses  can  each  man 
purchase?  Ans.  B  5,  C  9,  and  D  12. 

18.  How  many  rails  will  inclose  a  field  14599  feet  long  by 
10361  feet  wide,  provided  the  fence  is  straight,  and  7  rails  high, 
and  the  rails  of  equal  length,  and  the  longest  that  can  be  used  ? 

Ans.  26880. 

F 


82  PROPERTIES  OF  NUMBERS. 

LEAST  COMMON  MULTIPLE. 

t^l.    A  Multiple  is  a  number  exactly  divisible  by  a  given 

number;  thus,  20  is  a  multiple  of  4. 

Notes.  —  1.  A  multiple  is  necessarily  composite;  a  divisor  may  be  either 
prime  or  composite. 

2.  A  number  is  a  divisor  of  all  its  multiples  and  a  multiple  of  all  its  divisors. 

152.  A  Common  Multiple  is  a  number  exactly  divisible  by 
two  or  more  given  numbers ;  thus,  20  is  a  common  multiple  of  2, 
4,  5,  and  10. 

lo3.  The  Least  Common  Multiple  of  two  or  more  numbers 
is  the  least  number  exactly  divisible  by  those  numbers ;  thus,  24 
is  the  least  common  multiple  of  3,  4,  6,  and  8. 

154:.  From  the  definition  it  is  evident  that  the  product  of  two 
or  more  numbers,  or  any  number  of  times  their  product,  must  be 
a  common  multiple  of  the  numbers.  Hence,  A  common  multiple 
of  two  or  more  numhers  may  he  found  hy  multiplying  the  given 
numbers  together. 

155.   To  find  the  least  common  multiple. 

FIRST    METHOD. 

From  the  relations  of  multiple  and  divisor  we  have  the  following 
properties  : 

I.  A  multiple  of  a  number  must  contain  all  the  prime  factors 
of  that  number. 

II.  A  common  multiple  of  two  or  more  numbers  must  contain 
all  the  prime  factors  of  each  of  those  numbers. 

III.  The  least  common  multiple  of  two  or  more  numbers  must 
contain  all  the  prime  factors  of  each  of  those  numbers,  and  no. 
other  factors. 

1.  Find  the  least  common  multiple  of  63,  66,  and  78. 

OPERATION".  Analysis.      The 

63  =  3   X   3   X      7  number  cannot  be  less 

66  =  2  X  3  X  11  than   78,    because    it 

/  8  =  ^   X   ^   X   lo  must  contain  78  ;  and 

2x3x13x11x3x7  =  18018  Ans.      if  it   contains   78,   it 

must  contain   all   its 
prime  factors,  viz. ;         2  X  3  X  13. 


LEAST  COMMON  MULTIPLE.  33 

We  here  have  all  the  prime  factors,  and  also  all  the  factors  of  66 
except  11.     Annexing  11  to  the  series  of  factors, 

2  X  3  X  13  X  11, 
and  we  have  all  the  prime  factors  of  78  and  66,  and  also  all  the  fac- 
tors of  63  except  one  3,  and  7.     Annexing  3  and  7  to  the  series  of 

factors, 

2  X  3  X  13  X  11  X  3  X  7, 

and  we  have  all  the  prime  factors  of  each  of  the  given  numbers,  and 

no  others;  hence  the  product  of  this  series  of  factors  is  the  least 

common  multiple  of  the  given  numbers,  (III). 

From  this  example  and  analysis  we  deduce  the  following 

Rule.     I.   Resolve  the  given  numbers  into  their  prinie  factors. 

II.   Multiple/  together  all  the  prime  factors  of  the  largest  numher^ 

and  such  prime  factors  of  the  other  numbers  as  are  not  found  in 

the  largest  number j  and  their  product  will  be  the  least  common 

multiple. 

NoTR.  —  When  a  prime  factor  is  repented  in  any  of  the  given  numbers,  it 
must  be  taken  as  many  times  in  the  multiple,  as  the  greatest  number  of  times  it 
appears  in  any  of  the  given  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  Find  the  least  common  multiple  of  60^  84,  and  132. 

Alls.  4620. 

2.  Find  the  least  common  multiple  of  21,  80,  44,  and  126. 

Ans.  13,860. 

3.  Find  the  least  common  multiple  of  8,  12,  20,  and  30. 

4.  Find  the  least  common  multiple  of  16,  60,  140,  and  210. 

Ans.  1,680. 

5.  Find  the  least  common  multiple  of  7,  15,  21,  25,  and  35. 

6.  Find  the  least  common  multiple  of  14,  19,  38,  42,  and  57. 

Ans.  798. 

7.  Find  the  least  common  multiple  of  144,  240,  480,  960. 


SECOND    METHOD. 

1^6.   1.  What  is  the  least  common  multiple  of  4,  9,  12,  18, 

and  36  ? 


84  PROPERTIES  OF  NUMBERS. 


2 

4. 

OPERATION. 

.  9  .  .  12  .  .  18  . 

.  36 

2 

2  . 

.  9  .  .    6  .  .    9 

.  18 

3 

9.  .    3 . .    9 

.    9 

3 

3         3 

3 

2  X  2  X  3  X  3  = 


Analysis.  We  first  write 
the  given  numbers  in  a  se- 
ries with  a  vertical  line  at 
the  left.  Since  2  is  a  fac- 
tor of  some  of  the  given 
numbers,  it  must  be  a  factor 
of  the  least  common  mul- 
tiple sought,  (155,111).  Di- 


viding as  many  of  the  numbers  as  are  divisible  by  2,  we  write  the 
quotients,  and  the  undivided  number,  9,  in  a  line  underneath.  Now, 
since  some  of  the  numbers  in  the  second  line  contain  the  factor  2,  the 
least  common  multiple  must  contain  another  2,  and  we  again  divide 
by  2,  omitting  to  write  any  quotient  when  it  is  1.  We  next  divide 
by  3  for  a  like  reason,  and  still  again  by  3.  By  this  process  we  have 
transferred  all  the  factors  of  each  of  the  numbers  to  the  left  of  tho 
vertical ;  and  their  product,  3G,  must  be  the  least  common  multiple 
sought,  (155,  III). 

2.  What  is  the  least  common  multiple  of  20,  12,  15,  and  75  ? 

OPERATION.  Analysis.      We     readily 

see  that  2  and  5  are  among 
the  factors  of  the  given  num- 
bers, and  must  be  factors  of 
the  least  common  multiple  ; 
2x5x2x3x5  =  300,  Ans.      hence,  writing  2  and  5  at  the 

left,  we  divide  every  number 
that  is  divisible  by  either  of  these  factors  or  by  their  product ;  thus, 
we  divide  20  by  both  2  and  5  ;  12  by  2 ;  15  by  5  ;  and  75  by  5.  We 
next  divide  the  second  line  in  like  manner  by  2  and  3  ;  and  afterward 
the  third  line  by  5.  By  this  process  we  collect  the  factors  of  the 
given  numbers  into  groups ;  and  the  product  of  the  factors  at  the 
left  of  the  vertical  is  the  least  common  multiple  sought. 

3.  What  is  the  least  common  multiple  of  7, 10,  15,  42,  and  70? 

OPERATION.  Analysis.      In   this   operation 


2,5 

20. 

.12. 

.15. 

.75 

2,8 

2  . 

.    6. 

.    3  . 

.  15 

5 

5 

3,7 
2,5 


10 


15        4*^        70  ^'®  omit  the  7   and  10,  because 

they    are    exactly    contained    in 
some  of  the  other  given  numbers  ; 
3x7x2x5  =  210,  Arts.      thus,  7  is  contained  in  42,  and  10 

in  70  ;  and  whatever  will  contain 
42  and  70  must  contain  7  and  10.  Hence  we  have  only  to  find  the 
least  common  multiple  of  the  remaining  numbers,  15,  42,  and  70. 


LEAST  COMMON  MULTIPLE.  85 

From  these  examples  we  derive  the  following 

Rule.  I.  Write  the  numbers  in  a  Itnej  omitting  such  of  the 
smaller  numbers  as  ai^e  factors  of  the  larger ^  and  draw  a  vertical 
line  at  the  left. 
%  II.  Divide  by  any  prime  factor  or  factors  that  may  be  contained 
in  one  or  more  of  the  given  numbers j  and  write  the  quotients  and 
undivided  numbers  in  a  line  underneath,  omitting  the  Vs. 

III.  In  like  manner  divide  the  quotients  and  undivided  numberSy 
and  continue  the  process  till  all  the  factors  of  the  given  numbers 
have  been  transferred  to  the  left  of  the  vertical.  Then  multiply 
these  factors  together ^  and  their  product  will  be  the  l^ast  common 
multiple  required. 

Note.  —  We  may  use  a  composite  number  for  a  divisor,  when  it  is  contained 
in  all  the  given  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  least  common  multiple  of  15,  18,  21,  24,  35, 
36,  42,  50,  and  60  ?  Ans.  12600. 

2.  What  is  the  least  common  multiple  of  6,  8,  10,  15,  18,  20, 
and  24  ?  Ans.  360. 

3.  What  is  the  least  common  multiple  of  9,  15,  25,  35,  45,  and 
100?  Ans.  6300. 

4.  What  is  the  least  common  multiple  of  18,  27,  36,  and  40  ? 

5.  What  is  the  least  common  multiple  of  12,  26,  and  52  ? 

6.  What  is  the  least  common  multiple  of  32,  34,  and  36  ? 

Ans.  4896. 

7.  What  is  the  least  common  multiple  of  8, 12,  18,  24,  27,  and 
36? 

8.*  What  is  the  least  common  multiple  of  22,  33,  44,  55,  and 
66? 

9.  What  is  the  least  common  multiple  of  <34,  84,  96,  and  210  ? 

10.  If  A  can  build  14  rods  of  fence  in  a  day,  B  25  rods,  C  8 
rods,  and  D  20  rods,  what  is  the  least  number  of  rods  that  will 
furnish  a  number  of  whole  days'  work  to  either  one  of  the  four 
men?  Aiis.  1400. 

8 


go  PROPERTIES  OF  NUMBERS. 

11.  "What  is  the  smallest  sum  of  money  for  which  I  can  pur- 
chase either  sheep  at  $4  per  head,  or  cows  at  $21^  or  oxen  at  $49, 
or  horses  at  872?  Ans.  $3528. 

12.  A  can  dig  4  rods  of  ditch  in  a  day,  B  can  dig  8  rods,  and 
C  can  dig  6  rods ;  what  must  be  the  length  of  the  shortest  ditch, 
that  will  furnish  exact  days'  labor  either  for  each  working  alone 
or  for  all  working  together  ?  Ans.  72  rods. 

13.  The  foi-ward  wheel  of  a  carriage  was  11  feet  in  circumfer- 
ence, and  the  hind  wheel  15  feet;  a  rivet  in  the  tire  of  each  was 
up  when  the  carriage  started,  and  when  it  stopped  the  same  rivets 
were  up  together  for  the  575th  time;  how  many  miles  had  the 
carriage  traveled,  allowing  5280  feet  to  the  mile  ? 

Ans.  17  miles  5115  feet. 

CANCELLATION. 
157*    Cancellation  is  the  process  of  rejecting  equal  factors 
from  numbers  sustaining  to  each  other  the  relation  of  dividend 
and  divisor. 

158.  It  is  evident  that  factors  common  to  the  dividend  and 
divisor  may  be  rejected  without  changing  the  quotient,  (117, 
III). 

1.  Divide  1365  by  105. 

^T,^r>.rT,T^>.  Analysis.     We  first  in- 

OPERATION. 

-lo/-.-        ^        L        ^       -ir.  dicate  the  division  by  wri- 

= - . =  13  tii^g  the  dividend  above  a 

105  ?i  X  fi  Xt  horizontal  line  and  the  di- 

visor below.  Then  factor- 
ing each  term,  we  find  that  3,  5,  and  7  are  common  factors ;  and 
crossing,  or  canceling  these  factors,  we  have  13,  the  remaining  factor 
of  the  dividend,  for  a  quotient. 

151^.  If  the  product  of  several  numbers  is  to  be  divided  by 
the  product  of  several  other  numbers,  the  common  factors  should 
be  canceled  before  the  multiplications  are  performed,  for  two 
reasons : 

1st.  The  operations  in  multiplication  and  division  will  thus  be 
abridf^ed. 


CANCELLATION.  g^ 

2d.  The  factors  of  small  numbers  are  generally  more  readily 
detected  than  those  of  large  numbers. 
2.  Divide  20  times  66  by  7  times  15. 

OPERATION.  Analysis.     Having  first  indi- 

4        g  cated  all  the  operations  required 

^0  X  ^i       32  hy  the   question,   we    cancel  7 

~T        77  ^^    o~  ^^^^  from  7  and  5G,  and  5  from  15 

'^q  and  20,  leaving  the  factors  3  in 

the  divisor,  and  8  and  4  in  the 
dividend.  Then  8  X  4  =  32,  which  divided  by  3,  gives  10|,  the  quo- 
tient required.      Hence  the  following 

Rule  I.  Write  the  numbers  composing  the  dividend  above  a 
horizontal  line^  and  the  numbers  composing  the  divisor  below  it. 

II.  Cancel  all  the  factors  common  to  both  dividend  and  divisor. 

III.  Divide  the  product  of  the  remaining  factors  of  the  dividend 
by  the  product  of  the  remaining  factors  of  the  divisor^  and  the 
result  xvill  be  the  quotient. 

NoTKS.  —  1.  When  a  factor  is  canceled,  the  unit,  1,  is  supposed  to  take  its 
place. 

2.  By  many  it  is  thought  more  convenient  to  write  the  factors  of  the  dividend 
on  the  right  of  a  vertical  line,  and  the  factors  of  the  divisor  on  the  left. 


EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  quotient  of  18  X  6  x  4  X  42  divided  by  4  x  9 
X  3  X  7x  6? 

FIRST    OPERATION.  SECOND    OPERATION. 

'^^  X  0  X  4  X  M'       =4  i 

^x^x^x^x^  ^ 

t 

4,  Ans. 

2.  Divide  the  product  of  21  x  8  x  60  x  8  x  6  by  7  X  12  X  3 
X  8  X  3.  Ans.  80. 

3.  The  product  of  the  numbers  16,  5,  14,  40,  16,  60,  and  50, 
fs  to  be  divided  by  the  product  of  the  numbers  40,  24,  50,  20,  7, 
and  10;  what  is  the  quotient?  ^^s-  ^-- 


88  PROPERTIES  OF  NUMBERS.  ' 

4.  Divide  the  continued  product  of  12,  5,  183,  18,  and  70  by 
the  continued  product  of  3,  14,  9,  5,  20,  and  6. 

5.  If  213  X  84  X  190  X  264  be  divided  by  30  X  56  x  3G, 
what  will  be  the  quotient  ? 

6.  Multiply  64  by  7  times  31  and  divide  the  product  by  8 
times  56,  multiply  this  quotient  by  15  times  88  and  divide  the 
product  by  55,  multiply  this  quotient  by  13  and  divide  the  pro- 
duct by  4  times  6.  Ans.  403. 

7.  How  many  cords  of  wood  at  $4  a  cord,  must  be  given  for  3 
tons  of  hay  at  $12  a  ton  ? 

8.  How  many  firkins  of  butter,  each  containing  56  pounds,  at 
15  cents  a  pound,  must  be  given  for  8  barrels  of  sugar,  each  con- 
taining 195  pounds,  at  7  cents  a  pound  ?  Ans.  13. 

9  A  grocer  sold  16  boxes  of  soap,  each  containing  66  pounds 
at  9  cents  a  pound,  and  received  as  pay  99  barrels  of  potatoes, 
each  containing  3  bushels ;  how  much  were  the  potatoes  worth  a 
bushel  ? 

10.  A  farmer  exchanged  480  bushels  of  corn  worth  70  cents  a 
bushel,  for  an  equal  number  of  bushels  of  barley  worth  84  cents  a 
bushel,  and  oats  worth  56  cents  a  bushel;  how  many  bushels  of 
each  did  he  receive  ?  Ans.  240. 

11.  A  merchant  sold  to  a  farmer  two  kinds  of  cloth,  one  kind  at 
75  cents  a  yard,  and  the  other  at  90  cents,  selling  him  twice  as 
many  yards  of  the  first  kind  as  of  the  second.  He  received  as  pay 
132  pounds  of  butter  at  20  cents  a  pound ;  how  many  yards  of 
each  kind  of  cloth  did  he  sell  ? 

Ans.  22  yards  of  the  first,  and  11  yards  of  the  second. 

12.  A  man  took  six  loads  of  potatoes  to  market,  each  load  con- 
taining 20  bags,  and  each  bag  2  bushels.  He  sold  them  at  44 
cents  a  bushel,  and  received  in  payment  8  chests  of  tea,  each  con- 
taining 22  pounds ;  how  much  was  the  tea  worth  a  pound  ? 

Ans.  60  cents. 


DEFINITIONS,  NOTATION,  AND  NUMERATION.  89 

FKACTIONS. 

DEFINITIONS,  NOTATION,  AND  NUMERATION. 

160*  When  it  is  necessary  to  express  a  quantity  less  than  a 
unit,  we  may  regard  the  unit  as  divided  into  some  number  of  equal 
parts,  and  use  one  of  these  parts  as  a  new  unit  of  less  value  than 
the  unit  divided.  Thus,  if  a  yard,  considered  as  an  integral  unit, 
be  divided  into  4  equal  parts,  then  one,  two,  or  three  of  these 
parts  will  constitute  a  number  less  than  a  unit.  The  parts  of  a 
unit  thus  used  are  called  fractional  units  ;  and  the  numbers  formed 
from  t\iQm^  fractional  numbers.     Hence 

161,  A  Fractional  unit  is  one  of  the  equal  parts  of  an  inte- 
gral unit. 

16 ^«  A  Fraction  is  a  fractional  unit,  or  a  collection  of  frac- 
tional units. 

1@3.  Fractional  units  take  their  name,  and  their  value,  from 
the  number  of  parts  into  which  the  integral  unit  is  divided.    Thus, 

If  a  unit  be  divided  into  2  equal  parts,  one  of  the  parts  is 
called  one  half  If  a  unit  be  divided  into  3  equal  parts,  one  of  the 
parts  is  called  one  third.  If  a  unit  be  divided  into  4  equal  parts, 
one  of  the  parts  is  called  one  fourth. 

And  it  is  evident  that  one  third  is  less  in  value  than  one  half 
one  fourth  less  than  one  third,  and  so  on. 

IG^i:*  To  express  a  fraction  by  figures,  two  integers  are  re 
quired  j  one  to  denote  the  number  of  parts  into  which  the  inte- 
gral unit  is  divided,  the  other  to  denote  the  number  of  parts  taken, 
or  the  number  of  fractional  units  in  the  collection.  The  farmer 
is  written  below  a  horizontal  line,  the  latter  above.     Thus, 

One  fifth  is  written         | 


One  half  is  written         ^ 

One  third  "  \ 

Two  thirds       "  | 

One  fourth      ^^  | 

Two  fourths     "  f 

Three  fourths  "  | 
8* 


Two  fifths          "  I 

One  seventh      "  ^ 

Three  eighths  "  | 

Five  ninths       ^^  f 

Eight  tenths     "  i% 


90    ,  FRACTIONS. 

16^.  The  Denominator  of  a  fraction  is  tlie  mimber  below  tlie 
line. 

It  denominates  or  names  the  fractional  unit,  and  it  shows  how 
many  fractional  units  are  equal  to  an  integral  unit. 
'I@@.    The  Numerator  is  the  number  above  the  line. 

It  numerates  or  numbers  the  fractional  units ;  and  it  shows 
how  many  are  taken. 

167.  The  Terms  of  a  fraction  are  the  numerator  and  deno- 
minator, taken  together. 

168.  Since  the  denominator  of  a  fraction  shows  how  many 
fractional  units  in  the  numerator  are  equal  to  1  integral  unit,  it 
follows, 

I.  That  the  value  of  a  fraction  in  integral  units,  is  the  quo- 
tient of  the  numerator  divided  by  the  denominator. 

■  II.  That    fractions    indicate  division,  the    numerator  being  a 
dividend  and  the  denominator  a  divisor. 

mo.  To  analyze  a  fraction  is  to  designate  and  describe  its 
numerator  and  denominator.     Thus  ^  is  analyzed  as  follows  :  — 

7  is  the  denominator^  and  shows  that  the  units  expressed  by  the 
numerator  are  seventlis. 

5  is  the  numerator,  and  shows  that  5  sevenths  are  taken. 

5  and  7  are  the  terms  of  the  fraction  considered  as  an  expres- 
sion of  division,  5  being  the  dividend  and  7  the  divisor. 

EXAMPLES    EOR    PRACTICE. 


Express  the  following  fractions  by  figures  :  — 
1.   Four  ninths.  Ans. 


2.  Seven  Ji/fiy-sixths.  Aris.  -^^, 

3 .  Sixteen  forty-eigliths. 

4.  Ninety-five  one  hundred  seventy-ninths. 

5.  Five  hundred  thirty-six /oitr  hundredths. 

.  6.  One  thousand  eight  hundred  fifty-seven  7iine  thousand  Jive 
hundred  twenty-firsts. 

7.  Twenty-five  thousand  eighty-sevenths. 

8.  Thirty  ten  thousand  eighty-seconds. 

9.  One  hundred  one  ten  millionths. 


I 


DEFINITIONS,  NOTATION,  AND  NUMERATION.  91 

Read  and  analyze  the  following  fractions:  — 

10  4  .   7.17.   45_  .  J72_  .  _48_  .  _  8 1_  .  456 
■*-^'      9^  lH  ^     38^  J  00^  37  5^  1009^  7HG3^  537* 

11  2  0.  87.   95  .  48.   75  .  175.  4_3  6  .  _7  6  6 
J--*-'   4  ^  30  >  J  00  ^  '12  ^  4  37  ^    ^      >       60  ^  4  57"5* 

19   4  6  7.  5  3^.  10000.    7  5__.  5007 
^•^'    ^3(1^  ;^4§^    75   ;  l^OOO;  30  01" 

iq   150.  4  36.  137  85.  15  0072.  I_0000i 
^*^'     53  7^  972^  4  7  95(J^  4  7500  0^  2000(52* 

Fractions  are  distinguished  as  Proj)er  and  Improper, 

170.    A  Proper  Fraction  is  one  whose  numerator  is  less  than 

its  denominator;  its  value  is  less  than  the  unit  1. 

1*^1.    An  Improper  Fraction  is  one  whose  numerator  equals 

or  exceeds  its  denominator;  its  value  is  never  less  than  the  unit  1. 

Notes. — 1.  The  value  of  a  proper  fraction,  always  being  less  than  a  unit,  can 
only  bo  expressed  in  a  fractional  form  ,  hence,  its  name. 

2.  The  value  of  an  improper  fraction,  always  being  equal  to,  or  greater  than 
a  unit,  can  always  be  expressed  in  some  other  form;  hence  its  name. 

ly^.  A  Mixed  E'umber  is  a  number  expressed  by  an  integer 
and  a  fraction. 

ITS.  Since  fractions  indicate  division,  (ISS,  II),  all  changes 
in  the  iermH  of  a  fraction  will  affect  the  value  of  that  fraction  ac- 
cording to  the  laws  of  division;  and  we  have  only  to  modify  the 
language  of  the  General  Principles  of  Division,  by  substituting 
the  words  rmmerator^  denominator^  and  fraction^  or  value  of  the 
fraction,  for  the  words  dividend,  divisor^  and  quotient,  respectively, 
and  we  shall  have  the  followins; 

GENERAL    PRINCIPLES    OF   FRACTIONS. 

174.  Prin.  I.  MidAiphjing  tlie  numerator  multiplies  the 
fraction^  and  dioiding  the  numerator  divides  the  fraction. 

Prin.  II.  Midtiplijing  the  denominator  divides  the  fraction, 
and  dividing  the  denominator  TRultiplies  the  fraction. 

Prin.  III.  Multiplying  or  dividing  Loth  terms  of  the  fraction 
hy  the  same  number,  does  not  alter  the  value  of  the  fraction. 

175.  These  three  principles  may  be  embraced  in  one 

general   LAW. 

A  change  in  the  numerator  produces  a  like  change  in  the 
value  of  the  fraction  ;  hut  a  change  in  the  denominator  j)ro(i?^cej 
an  OPPOSITE  change  in  the  value  of  the  fraction. 


OPERATION. 

l%% 

12    __ 

2  1 

Or, 

4 

1 

5)tVo  = 

4 

92  FRACTIONS. 

REDUCTION. 
"B  70.    The  Reduction  of  a  fraction  is  the  process  of  changing 
its  terms,  or  its  form,  without  altering  its  value. 

CASE   I. 

177.   To  reduce  fractions  to  their  lowest  terms. 

A  fraction  is  in  its  loivest  terms  when  its  numerator  and  denomi- 
nator are  prime  to  each  other ;  that  is,  when  both  terms  have  no 
common  divisor. 

I.  Eeduce  the  fraction  y^^^  to  its  lowest  terms. 

Analysis.    Dividing  both  terms  of 

the  fraction  by  the  same  number  does 
not  alter  the  value  of  the  fraction, 
(174,  III);  hence,  we  divide  both 
terms  of  yg^  by  5,  and  both  terms  of 
the  result,  ^f ,  by  3,  and  obtain  4  for  the  final  result.  As  4  and  7  are 
prime  to  each  other,  the  lowest  terms  of  j^^  are  ^. 

Instead  of  dividing  by  the  factors  5  and  3  successively,  we  may 
divide  by  the  greatest  common  divisor  of  the  given  terms,  and  reduce 
the  fraction  to  its  lowest  terms  at  a  single  operation.     Hence,  the 

Rule.  Cancel  or  reject  all  factors  common  to  hofh  numerator 
and  denominator.      Or, 

Divide  both  terms  hy  their  greatest  common  divisor, 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  j'^^^^j  to  its  lowest  terms.  Ans.  | 

2.  Reduce  f|  to  its  lowest  terms.  Ans.  |. 
8.  Reduce  H|  to  its  lowest  terms.                             Ans.  |. 

4.  Reduce  -^^^  to  its  lowest  terms.  Ans.  |. 

5.  Reduce  ^if  ^^  i^s  lowest  terms.  Ans.  |. 

6.  Reduce  ||f  to  its  lowest  terms. 

7.  Reduce  |||  to  its  lowest  terms. 

8.  Reduce  j^^.^^  to  its  lowest  terms. 

9.  Reduce  f  §||  to  its  lowest  terms. 

10.  Reduce  |^|  to  its  lowest  terms.  Ans.  -||. 

Note.  —  Consult  the  factor  table. 

II.  Reduce  j^Y?  *^  ^^^  lowest  terms.  Ans.  ||. 


REDUCTION.  93 

12.  Reduce  f^fl  to  its  lowest  terms.  Ans.   |f. 

13.  Reduce  |||f  to  its  lowest  terms. 

14.  Reduce  || J^  to  its  lowest  terms.  Ans. 

15.  Reduce  |i|||,  and  ^||f  g  to  their  lowest  terms. 


4  1 


CASE    II. 

17§.   To  reduce  an  improper  fraction  to  a  whole  or 
mixed  number. 

1.  Reduce  ^-f^  to  a  whole  or  mixed  number. 

OPERATION. 

2_9J7  =  297  -^  12  =  24-^-c  =  24^ 

Analysis.     Since  the  value  of  a  fraction  in  integral  units  is  equal 

to  the  quotient  of  the  numerator  divided  by  the  denominator,  (168,  1,) 

we  divide  the  given  numerator,  297,  by  the  given  denominator,  12, 

and  obtain  for  the  value  of  the  fraction,  the  mixed  number  24y'2  =  24f . 

Hence  the 

Rule.     Divide  the  numerator  hy  the  denominator. 

Notes.     1.  When  the  denominator  is  an  exact  divisor  of  the  numerator,  the 
result  will  be  a  whole  number. 

2.  In   all  answers  containing  fractions,  reduce  tbo  fractions  to  their  lowest 
terms. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  ^^  to  an  equivalent  integer.  Ans.  16. 

2.  Reduce  ^-^  to  an  equivalent  integer. 

3.  Reduce  ^g"*  to  a  mixed  number.  Ans.  17^. 

4.  Reduce  ^^^^  to  a  mixed  number.  Ans.  26||, 

5.  Reduce  ^^^  to  a  mixed  number,  Ans.  24|. 

6.  Reduce  ^g®  to  a  mixed  number.  Ans.  17^|. 

7.  Reduce  ^||*  to  a  mixed  number. 

8.  Reduce  ^  j^|^  to  a  mixed  number,  Ans.  156|. 

9.  Reduce  3|02  ^q  ^  mixed  number. 

10.  Reduce  g^/g®  to  a  mixed  number.  Ans.  4|J. 

11.  Reduce  ^fff^  to  a  mixed  number.  Ans.  100|. 

12.  Reduce  j^l^f  to  a  mixed  number. 

13.  In  '^^^^  of  a  day  how  many  (Jays?         Ans.  982|  days. 

14.  In  ^f^  of  a  dollar  how  many  dollars?  Ans.  %^\^^. 

15.  If  1000  dollars  be  distributed  equally  among  36  men,  what 
part  of  a  dollar  must  each  man  receive  in  change  ?        Ans.   J. 


94  FRACTIONS. 

CASE  III. 

179.  To  reduce  a  whole  number  to  a  fraction  having 
a  given  denominator. 

1.   lleduce  37  to  an  equivalent  fraction  whose  denominator  shall 

be  5. 

OPERATION.  Analysis.    Since  in  each  unit  there  are 

S7  X  ^  =  1S5  ^  fifths,  in  37  units  there  must  be  37  times 

37  ^  185^  j^^.  5  fifths,  or  185  fifths  =  n^ .     The  nume- 
rator, 185,  is  obtained  in  the  operation  by 

multiplying  the  whole  number,  37,  by  the  given   denominator,  5. 
Hence  the 

HuLE.  3fultipli/  the  whole  number  hy  the  given  denominator  ; 
talce  the  product  for  a  number atorj  under  which  write  the  given  de- 
nominator. 

Note. — A  whole  number  may  be  rerluced  to  a  fractional  form  by  writing  1 
under  it  for  a  denominator;  thus,  9  =  ^-. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  17  to  an  equivalent  fraction  whose  denominator 
shall  be  6.  Ans,    ^g^. 

2.  Change  375  to  a  fraction  whose  denominator  shall  be  8. 

3.  Change  478  to  a  fraction  whose  denominator  shall  be  24. 

4.  Reduce  36  pounds  to  ninths  of.  a  pound. 

5.  Reduce  359  days  to  sevenths  of  a  day.  Ans^.  ^y  ^ 

6.  Reduce  763  feet  to  fourteenths  of  a  foot.        Ans.   ^^y\®^. 

7.  Reduce  937  to  a  fractional  form.  Ans.   ^^^ . 

CASE   IV. 

180.  To  reduce  a  mixed  number  to  an  improper  frac- 
tion. 

1.  In  12|  how  many  sevenths? 

OPERATION.  Analysis.     In  the  whole  number  12,  there  are 

1-7  12  X  7  sevenths  =  84  sevenths,  (Case  III),   and 

'  84  sevenths  -f  5  sevenths  =:  89  sevenths,  or  \^. 

89  Hence  the  following 


I 


REDUCTION.  95 


Rule.  Multiply  the  whole  number  hy  the  denominator  of  the 
fraction ;  to  the  product  add  the  numerator,  and  under  the  sum 
write  the  denominator. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  154  to  fifths.  Ans.  ^f. 

2.  Reduce  24|  to  an  improper  fraction.  Ans.   ^■^. 

3.  Reduce  57 f  to  an  improper  fraction. 

4.  Reduce  356i|  to  an  improper  fraction.  Ans.   ^^|^. 

5.  Reduce  872^^3^  to  an  improper  fraction. 

G.  Reduce  800 ^^^  to  an  improper  fraction.  ^    Ans.    ^|^§^. 

7.  Reduce  434^|  to  an  improper  fraction.  Ans.   ^%^3^^. 

8.  In  15|  how  many  eighths? 

9.  In  135^%  how  many  twentieths?  Ans.  ^-J^^. 

10.  In  43|  bushels  how  many  fourths  of  a  bushel  ? 

11.  In  760 j^^  days  how  many  tenths  of  a  day? 

CASE   V. 

181.    To  reduce  a  fraction  to  a  given  denominator. 
We  have  seen  that  fractions  may  be  reduced  to  lower  terms  by 
division.     Conversely, 

I.  Fractions  may  be  reduced  to  higher  terms  by  multiplication. 

II.  All  higher  terms  of  a  fraction  must  be  multiples  of  its 
lowest  terms. 

1.  Reduce  -|  to  a  fraction  whose  denominator  is  40. 

OPERATION.  Analysis.     We  first  divide  40,  the  re- 

40  -f-  8  =  5  quired  denominator,  by  8,  the  denomi- 

3^5  nator  of  the  given  fraction,  to  ascertain 

8  X  5  "^  ^^'  ^^^         ^^  ^*  ^^  ^  multiple  of  this  term,  8.     The 
division  shows  that  it  is  a  multiple,  and 
that  5  is  the  factor  which  must  be  employed  to  produce  it.  We  there- 
fore multiply  both  terms  of  |  by  5,  (174,  III),  and  obtain  {§,  the  re- 
quired result.     Hence  the 

Rule.  Divide  the  required  denominator  hy  the  denominator 
of  the  given  fraction^  and  multiply  both  terms  of  the  fraction  hy 
the  quotient. 


96  rrtACTioNS. 


EXAMPLES   FOR   PRACTICE. 
1.  Eeduce  |  to  a  fraction  having  24  for  a  denominator. 


Alls.  ^|. 


2.  Reduce  -^^  to  a  fraction  whose  denominator  is  96. 

Ans.  f|. 

3.  Reduce  |f  to  a  fraction  whose  denominator  is  51. 

4.  Reduce  y^g  to  a  fraction  whose  denominator  is  78. 

6.  Reduce  gW  to  a  fraction  whose  denominator  is  3000. 

A'ijv       4  9  6 

6.  Change  7|  to  a  fraction  whose  denominator  is  8. 

7.  Change  IOt/^  to  a  fraction  whose  denominator  is  176. 

8.  Change  bj\  to  a  fraction  whose  denominator  is  363. 

9.  Change  36f  to  a  fraction  whose  denominator  is  42. 

Ans.   If 42. 

CASE   VI. 

18S8.     To  reduce  two  or  more  fractions  to  a  common 
denominator. 

A  Common  Denominator  is  a  denominator  common  to  two  or 
more  fractions. 

1.  Reduce  |  and  |  to  a  common  denominator. 

Analysis.     We  multiply  the  terms  of  the 
OPERATION.  £j,g^  fraction  by  the  denominator  of  the  second, 

^  ^  "—=27  and  the  terms  of  the  second  fraction  by  the 

5x9        ^^  denominator  of   the   first,    (174,    III).     This 

Y  v>  5  must  reduce  each  fraction  to  the  same  deno- 

o"  w   5  ^^  4  o  rainator,  for  each  new  denominator  will  be  the 

product  of  the  given  denominators.   Hence  the 

Rule.     Multiply/  the  terms  of  each  fraction  hy  the  denominators 
of  all  the  other  fractions. 

Note. — Mixed  numbers  must  first  be  reduced  to  improper  fractions. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  |  and  |  to  a  common  denominator.     Ans.  H,  -j-^. 

2.  Reduce  ^  and  |  to  a  common  denominator. 


REDUCTION.  97 

3.  Reduce  |,  -f^  and  i   to  a  common  denominator. 

JtJS       "72         50        60 

ji.ns.  -j5^,  j^^y  ^2^. 

4.  Reduce  ^,  5|  and  1|  to  equivalent  fractions  having  a  com- 
mon denominator.  Ans.  -1|,  y^j^,  ||. 

5.  Reduce  y^^  and  j^^  to  a  common  denominator. 

Ay-,^        6  8        3  9 
^^^5-    22T,  -z^j- 

6.  Reduce  ^,  ^  and  Jj  to  a  common  denominator. 

7.  Reduce  |,  |,  j'^2  ^"^  /s  ^^  ^  common  denominator. 

/j„jf        768       115  2       896         864 
•^"^-     To  3  5?  153  5?  75  3  5?  T5^S* 

CASE    VII. 

183.  To  reduce  fractions  to  their  least  •common  de- 
nominator. 

The  Least  Common  Denominator  of  two  or  more  fractions  is 
the  least  denominator  to  which  they  can  all  be  reduced. 

184,  We  have  seen  that  all  higher  terms  of  a  fraction  must 
be  multiples  of  its  lowest  terms,  (181,  II)-     Hence, 

I.  If  two  or  more  fractions  be  reduced  to  a  common  denomi- 
nator, this  common  denominator  will  be  a  common  multiple  of  the 
several  denominators. 

II.  The  least  common  denominator  must  therefore  be  the  least 
common  multiple  of  the  several  denominators. 

1.  Reduce  |,  ^^  and  f^  to  their  least  common  denominator. 

OPERATION.  Analysis.     We  first  find  the  least 

12  . .  15  common  multiple  of  the  given  deno- 

.  minators,  which  is  60.     This  must 
4 
be  the  least  common  denominator  to 


3    ,   5 
2    ,   2 


3x5x2x2  =  60  which  the  given  fractions  can  be  re- 

duced, (II).  Reducing  each  frac- 
tion to  the  denominator  60,  by  Case 
V,  we  obtain  f  J,  ||  and  /„  ^^r  the 
answer.     Hence  the  following 

Rule.     I.  Find  the  least  common  multiple  of  the  given  denoTti" 
inators,  for  the  least  common  denominator. 
9  G 


98  FRACTIONS. 

II.  Divide  this  common  denominator  hy  each  of  the  given  d&- 
nominators^  and  multipli/  each  numerator  hi/  the  corresponding 
quotient.      The  products  loill  be  the  neic  numerators. 

Notes. — 1.  If  the  several  fractions  are  not  in  their  lowest  terms,  they  should 
be  reduced  to  their  lowest  terms  before  applying  the  rule. 

2.  When  two  or  more  fractions  are  reduced  to  their  least  common  denominator, 
their  numerators  and  the  cummon  denominator  will  be  prime  to  each  other. 

EXAMPLES    FOR   PRACTICE. 

1       1.  Reduce  |  and  f^  to  their  least  common  denominator. 

/<  ^  Q     2  5      12 

2.  Reduce  |,  |  and  |  to  their  least  common  denominator. 
8.  Reduce*  |,  -^^  and  |^  to  their  least  common  denominator. 

4.  Reduce  |,  |  and  |  to  their  least  common  denominator. 

5.  Reduce  j^^,  i|  and  ||  to  their  least  common  denominator. 

Jjjd     3  6      3  5      2  6 

6.  Reduce  |,  7*3?  ||  and  -^-^  to  their  least  common  denominator. 


J.JO     5  2     2  4      7  5       8 


7.  Reduce  2|,  ^-^^  ^^  and  |g  to  their  least  common  denomi- 

v.qfnr  Aii^i      312        5  6        25         74 

l^^^OX.  imS.     y2^,     J33^,     y2^,     JTTQ. 

8.  Reduce  |^,  g^g  and  |J  to  their  least  common  denominator. 

9.  Reduce  |g,  ^-^j^  and  ^|  to  their  least  common  denominator- 

A..<i   60    20    21 

10.  Reduce  ^-^^^  -jY^  and  ;||S  to  their  least  common  denomi- 
nator. Ans,  ^^,  yf,  £-^^j. 

11.  Reduce  l|^,  ||^  and  j-glj  to  their  least  common  denomi- 

■nnfnr  An<i      371         901         713 

naior.  ^?t.s.    j-^y-g,    i-:2-i^?  "121^* 

12.  Reduce  2|,  j^^  and  1-j^^  to  their  least  common  denominator. 

13.  Reduce  jgV^?  1411  ^^^  fell  to  their  least  common  denom 

inof  nr  4  ix      18  0  6  14      19  3  6  7  4      3  2  3  7  :?  3 

l'^^^*^^-  ■^^'^^-    3o4  8ig'   3  518  2  5'  '3^>.|6•i^r 

14.  Reduce  -|,   {A,  f.,   ^s^,  /^  and  |g  to  their  least  common 

flpnmninntnr  An<i      .'>400     6  930      1008     2240      1944     32i_3 

aenominaior.  jins.  ^^qq,  ^ggo'  7ogo'  7  5G0'  7  56T5?  7560 

15.  Reduce  ^,  A,  rp^j  ^^  and   j'g^^  to  their  least  common  de- 
nominator. 

16.  Reduce  j\,  t^\,  ||  and  4^  to  their  least  common   denomi- 

r^o+r»T.  Aiif:       2  8  7  6  0       45  5 

IiatOr.  .^?IS.    y^^,   y^5,    JXJ-,   y^jj. 


ADDITION.  99 

ADDITION. 

185.  The  denominator  of  a  fraction  determines  the  value  of 
the  fractional  unit,  (16o)  ;  hence, 

I.  If  two  or  more  fractions  have  the  same  denominator,  their 
numerators  express  fractional  units  of  the  same  value. 

II.  If  two  or  more  fractions  have  different  denominators,  their 
numerators  express  fractional  units  of  different  values. 

And  since  units  of  the  same  value  only  can  be  united  into  one 
sum,  it  follows, 

III.  That  fractions  can  be  added  only  when  they  have  a  com- 
mon denominator. 

1.  What  is  the  sum  of  4,  j%  and   ^  ? 

O^     1 ^  JO 

OPERATION. 

«  +   «   +   2  _  12  +  25  +  8  _  ,,,  _  3  . 

5     '     T3Z     '     1  5  —  0Q  —  gU  —  ?• 

Analysis.  We  first  reduce  the  given  fractions  to  a  common  deno- 
minator, (III).  And  as  the  resulting  fractions,  J§,  |^,  and  /j  have  the 
same  fractional  unit,  (I),  we  add  them  by  uniting  their  numerators 
into  one  sum,  making  ||  —  J,  the  answer. 

2.  Add  5|,  3|  and  4/3. 

Analysis.      The    sum   of   the 

OPERATION.  integers,  5,  3,  and  4,  is  12;  the 

^  +  5  +    i^  =  1^  sum  of  the  fractions,  J,  J,  and 

34_7_| 7      95  '4'K' 

^  ^  H    J^  ]2  —     ^2i  -7-    is  24.     Hence,  the  sum  of 

14^''^,  Ans.       both   fractions    and    integers   is 
12  +  2^^  =  14^^. 
180,    From  these  principles  and  illustrations  we   derive  the 
following  general 

Rule.  I.  To  add  fractions. —  When  necessary,  reduce  tlie.frac' 
tioiis  to  their  least  common  cJenominator ;  then  add  the  numerators 
and  place  the  sum  over  the  common  denominator. 

II.  To  add  mixed  numbers.  —  Add  the  integers  and  fractions 
separately^  and  then  add  their  sums. 

Note. — All  fractional  results  should  be  reduced  to  their  lowest  termS;  and  if 
improper  fractions,  to  whole  or  mixed  numbers. 


100  FRACTIONS. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  sum  of  Z^,  y^^,  /^  and  ^  ?  Ans.  2\, 

2.  What  is  the  sum  of  ||,  j\y  f^  and  f^  ?  Ans.  1|. 

3.  What  is  the  sum  of  /j,  r^-^y  ^f  and  ^f  ? 

4.  What  is  the  sum  of  1^,  Sfl,  2||,  5^§  and  4||  ? 

^rw.  28|. 

5.  What  is  the  sum  of  S7^%,  12|5,  13||  and  f  |  ? 

6.  Add  I,  I  and  |. 


7.  Add  I,  I,  If  and  j-\.  J.n«.  2 


^^• 


8.  Add  1,  I  and  J3. 

9.  Add  ,%  II  and  /^.  ^n^.  l|g. 

10.  Add  1,  f  1^  and  f  |.  jiws.  2||. 

11.  Add  I,  1^,11,11  and  if  ^n.4/,V 

12.  Add  3^,  4|  and  2/3. 

13.  Add  16^2  and  24  Jg.  Ans.  40/^. 

14.  Add  1^,  2f ,  3|,  4|  and  5|. 

15.  Add  4/3,  82^  and  2/^.  Ans.  14if 

16.  Add  I,  |,  jKj  ^°^  T7-  -^^-  It- 

17.  Add  ^,  f ,  j2^  and  -\. 

18.  Add  1,  I,  y\,  3\  and  ^|.  Ans.  Iff. 

19.  Add-i,  j^V/gandf. 

20.  Add  41i,  105|,  300|,  241|  and  472f        Ans.  1161|g. 

21.  Add  4^,  2|,  1 J^,  2/^,  5/5,  7|,  4»  and  6|. 

22.  Four  cheeses  weighed  respectively  36|,  42|,  30/g  and  51| 
pounds;  what  was  their  entire  weight?         Anfs.  169||  pounds. 

23.  What  number  is  that  from  which  if  4|  be  taken,  the  re- 
mainder will  be  3||?  Ans.  8|. 

24.  What  fraction  is  that  which  exceeds  /g  by  ^-f  ? 

25.  A  beggar  obtained  |  of  a  dollar  from  one  person,  J  from 
another,  ^  from  another,  and  f^  from  another;  how  much  did  he 
get  from  all  ? 

26.  A  merchant  sold  46|  yards  of  cloth  for  ^127/5,  64-^4  yards 
for  $226|,  and  76|  yards  for  831 2§  ;  how  many  yards  of  cloth  did 
he  sell,  and  how  much  did  he  receive  for  the  whole  ? 

Ans.  187|J  yards,  for  ?666j|. 


SUBTRACTION.  201 

SUBTRACTION. 

187.  The  process  of  subtracting  one  fraction  from  another  is 
based  upon  the  following  principles : 

I.  One  number  can  be  subtracted  from  another  only  when  the 
two  numbers  hare  the  same  unit  value.     Hence, 

II,  In  subtraction  of  fractions,  the  minuend  and  subtrahend 
must  have  a  common  denominator^  (ISSj  I). 

1.  From  I  subtract  §. 

OPERATION'.  Analysis.      Reducing   the 

4 2  __  1  2_r^i  0  __  _2_  given  fractions  to  a  common 

denominator,  the  resulting 
fractions  jf  and  {§  express  fractional  units  of  the  same  value,  (185, 
I).  Then  12  fifteenths  less  10  fifteenths  equals  2  fifteenths  =  j%,  the 
answer. 

2.  From  238^  take  24|. 

OPERATION.  Analysis.     We  first  reduce  the  frac- 

23 §1  3:;  238-^.  tional  parts,  j  and  |,  to  the  common 

9_j^5  --_     ^410  denominator,    12.      Since   we   cannot 

-1^/2  -'^^•-^-  making  ;|.  Then,  ||  subtracted  from 
f  f  leaves  f\  ;  and  carrying  1  to  24,  the  integral  part  of  the  subtrahend, 
(73,  II),  and  subtracting,  we  have  213y\  for  the  entire  remainder. 

ISS.  From  these  principles  and  illustrations  we  derive  the 
following  general 

EuLE.  I.  To  subtract  fractions. —  WJien  necessar^y  reduce  the 
frai'tions  to  tJietr  kast  cojiunon  deno7nwafor.  Subtract  (he  nume- 
rator  of  the  subtrahend  from  the  numerator  of  the  minuendy  and 
phice  the  difference  of  the  new  numerators  over  the  cominon  de^wm- 
I'nafor. 

II.  To  subtract  mixed  numbers.  —  Reduce  the  /raetional  parts 

to  a  co7n7non  denoininatory  and  then  subtract  the  fractional  and 

vite</ral  par^i  srj)rtra/f7y. 

NoTK. — We  may  reduce  mixed  immbers  to  improper  fractions,  and  subtract 
by  the  rule  for  fractions.  But  this  method  jrenerally  imposes  the  useless  labor 
of  reducing  integral  numbers  to  fractions,  and  fractions  to  integers  again. 

9* 


02                                              FRACTIONS. 

EXAMPLES   FOR   PRACTICE. 

1,   From  -j^g  take  /_. 

Arts.  j%. 

2.  From  l^  take  U. 

Ans.  |. 

3.  From  |f  take  ^-^, 

4.  From  |  take  |. 

Ans.  1|. 

5    From  1  take  ^-q. 

6.   From  ]|  take  /^. 

Arts.  ||. 

7    From  j\  take  i|. 

Ans.  j\. 

8.  From  J|  take  Jf. 

Ans.  J5. 

9.  From  /^  take  j\. 

10.   From  -^^^  take  /^. 

Ans.  13. 

11.   From  g\  take  jL^^. 

Ans.  yl-Q. 

12.  From  /^  take  /^^ 

13.  From  16|  take  7^- 

Ans.  9f . 

14.  From  36i|  take  8l|. 

Ans.  27|. 

15.  From  25/^  subtract  141|. 

16.  From  75  subtract  4|. 

Ans.  704. 

17.  From  18|  subtract  5|. 

18.  From  26^\  subtract  251|. 

19-  From  28i|  subtract  3-^94." 

Ans.  24if 

20.  From  78/.  subtract  32f . 

21.  The  sum  of  two  numbers  is  26|,  and  the  less  is  7y'3  ;  what 

is  the  greater  ? 

Ans.  19/^. 

22.  What  nmnber  is  that  to  which  if 

you 

add 

184,  the  sum 

wiUbe97f? 

23.  What  number  must  you  add  to  the 

sum 

of  1264  a«^  240f, 

to  make  560t  ? 

Ans.  193|i. 

24..  What  number  is  that  which^  added  to  the 

sum  of  ^,  J3J, 

and  j^^,  will  make  H? 

Ans.  1^. 

25.  To  what  fraction  must  |  be  added,  that  the  sum  may  be  |  ? 

26  From  a  barrel  of  vinegar  containing  31 J  gallons,  14  J  gallons 
were  drawn  ;  how  much  was  then  left?  Ans.  16|  gallons. 

27.  llought  a  quantity  of  coal  for  $140f,  and  of  lumber  for 
$456S.  Sold  the  coal  for  $775J,  and  the  lumber  for  $516y3g ;  how 
much  was  my  whole  gain  ?  Ans.  $694||. 


r 

l^K   THEORY  ' 


THEORY  OF  MULTIPLICATION  AND  DIVISION.  103 


THEORY  OF  MULTIPLICATION  AND  DIVISION  OF  FRACTIONS. 

ISO.    In  multiplication  and  division  of  fractions,  the  various 
operations  may  be  considered  in  two  classes : 
1st.  Multiplying  or  dividing  a  fraction* 
2d.  Multiplying  or  dividing  hy  a  fraction. 

100,  The  methods  of  multiplying  and  dividing  fractions  may 
be  derived  directly  from  the  General  Principles  of  Fractions, 
(1T4:);  as  follows : 

I.  To  multiply  a  fraction.— J/ii^^/j^/y  its  numerator  or  divide  its 
denominator  J  (tT4:,  I.  ond  IT). 

II.  To  divide  a  fraction. — Divide  its  numerator  or  muUipIi/  its 
denominator  J  (ITi,  I.  and  II). 

GENERAL   LAW. 

III.  Perfornfi  the  required  operation  upon  tlie  numerator^  or  the 
opposite  upon  the  denominator,  (IT^,  III). 

101.  The  methods  of  multiplying  and  dividing  hy  a  fraction 
may  be  deduced  as  follows : 

1st.  The  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator  (108,  I)-     IlencC; 

2d.  The  numerator  alone  is  as  many  times  the  value  of  the 
fraction,  as  there  are  units  in  the  denominator. 

3d.  If,  therefore,  in  multiplying  by  a  fraction,  we  multiply  by 
the  numerator,  this  result  will  be  too  great,  and  must  be  divided 
by  the  denominator. 

4th.  But  if  in  dividing  by  a  fraction,  we  divide  by  the  nume- 
rator, the  resulting  quotient  will  be  too  small,  and  must  be  multi- 
plied by  the  denominator. 

Hence,  the  methods  of  multiplying  and  dividing  hy  a;  fraction 
may  be  stated  as  follows : 

I.  To  multiply  by  a  fraction.  —  Multiply  hy  the  numerator  and 
divide  hy  the  denominator ,  (3d). 

II.  To  divide  by  a  fraction. — Divide  hy  the  numerator  and  mul- 
tiply hy  the  denominator,  (4th). 


104 


FRACTIONS. 


GENERAL   LAW. 
III.  Perform  the  required  operation  hy  the  numerator  and  the 
opposite  hy  the  denominator. 


MULTIPLICATION. 


193.   1.  Multiply  -^^  by  4. 


FIRST    OPERATION. 

5      v4.  20  12 

T3   X   ^  —  J2  ^z 


SECOND    OPERATION 

x4  =  | 


6 


12 


Analysis.  In  the  first  opera- 
tion, we  multiply  the  fraction 
by  4  by  multiplying  its  nume- 
rator by  4;  and  in  the  second 
operation,  we  multiply  the  frac- 
tion by  4  by  dividing  its  denom- 
inator by  4,  (190,  I  or  III). 

In  the  third  operation,  we  ex- 
press the  multiplier  in  the  form 
of  a  fraction,  indicate  the  mul- 
tiplication, and  obtain  the  result  by  cancellation. 


third  operation. 
3 


2.  Multiply  21  by  4. 


Analysis.  To  multiply  by  4» 
we  must  multiply  by  4  and  di- 
vide by  7,  (191,  I  or  III). 

In  the  first  operation,  we  first 
multiply  21  by  4,  and  then  di- 
vide the  product,  84,  by  7. 

In  the  second  operation,  we 
first  divide  21  by  7,  and  then 
multiply  the  quotient,  3,  by  4. 

In  the  third  operation,  we  ex- 
press the  whole  number,  21,  in 
the  form  of  a  fraction,  indicate  the  multiplication,  and  obtain  the 
result  by  cancellation. 


FIRST    operation. 

21  X  4  =  V  =  12 

SECOND    OPERATION. 

21x4  =  3x4  =  12 

THIRD    OPERATION. 

3 

i]i       4 


\^  t 


12 


3.  Multiply  j\  by  |. 


FIRST    OPERATION. 

Ist  Step, 

t\ 

X  7  = 

fl 

2d  step, 

II 

-8  = 

ih 

t¥5  = 

-h 

Am. 

ANALYsrs.  To  multiply  by 
|,  we  must  multiply  by  7  and 
divide  by  8,  (191,  I  or  III). 
In  the  first  operation,  we  muL 
tiply  /j-  by  7  and  obtain  f  | ; 


MULTIPLICATION. 


105 


SECOND  oPERATiox.  we  then  divide  ?|  by  8  and  obtain 

j5^  X  I  =  j^j\  =  j%  tVi*  which  reduced   gives   j\,  the 

required    product.     In   the    second 

THIRD    OPERATION.  ,.  i  i.    •      xi  ix 

operation  we  obtain  the  same  result 

^  '___  _5_  by  multiplying  the  numerators  to- 

At*        ^  gether  for  the  numerator  of  the  pro- 

-•  duct,  and  the  denominators  together 

for  the  denominator  of  the  product.    In  the  third  operation,  v^q  indicate 
the  multiplication,  and  obtain  the  result  by  cancellation. 

tl^3«  From  these  principle&  and  illustrations  we  derive  the 
followino;  sfcneral 

lluLE.  I.  Reduce  all  integers  and  mixed  numbers  to  improper 
fractions. 

II.  Multiply  together  the  mimerators  for  a  new  numerator ,  and 

the  denominators  for  a  new  denominator. 

Notes. — 1.  Cancel  all  factors  common  to  numerators  and  denominators. 
2.  If  a  fraction   be  multiplied   by   its   denominator,   the  product  will   be  t>>e 
numerator. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  I  by  8.  Ans.  2| 

2.  Multiply  I  by  27,  f^  by  4,  and  3^^  by  9. 


3.  Multiply  ^\  by  15.  Ans.  |. 

4.  Multiply  8  by  |.  Ans.  6. 

5.  Multiply  75  by  f-.,  7  by  ^fj,  756  by  |,  and   572  by  ^\, 

6.  Multiply  I  by  |. 

7.  Multiply  11  by  II,  and  jf  by  f  ^. 

8.  Multiply /3,- by  11,  and/,  by  i^. 

9.  Multiply  24  by  3f. 

10.  Multiply  If  by  1]|. 

11.  Multiply  3%  by  21 1. 

12.  Find  the  value  of  |  X  |  X  y\  X  |. 

13.  Find  the  value  of  |   X  |  X  -Jf  X  ^\  X  4|. 

14.  Find  the  value  of  ||  X   ^'^3  X  fff. 

15.  Find  the  value  of  2|  x  24  X  fj  X  y-gg  X  1/^  X  26}. 

Ans.  2. 

16.  Find  the  value  of  y\  x  -ij  X  4|  x  15  X  /g. 

17.  Find  the  value  of  ^^^^  x  ^izj  X  V/-  ^'^^'  lU- 


Ans, 

10 

Ans. 

91 

Ans. 

7 
34 

Ans. 

1 

X06  FRACTIONS. 


18.  Find  the  value  of  (4-^^  X  |)  +  If  X  (H  —  j%)- 

19.  Find  the  value  of  28^+  (7|  —  2|)  X  §  X  (f  +  4). 

KoTE  3.  —  The  word  of  between  fractions  is  equivalent  to  the  sign  of  multi- 
plication;   and   such   an   expression   is     sometimes  culled  ncotnpouud fraction. 

Find  the  values  of  the  following  indicated  products :  — 

20.  4  of  I  of  |.  Ans.  f . 

21.  f  of  I  of  3,3_.  Ans.   ^\. 

22.  f  of  j\  of  f . 

23.  1^  of  ^%  of  Jf  of  f|.  Alls.  Z^. 

24.  i  of  I  of  I  of  I  of  I  of  4  of  J  of  I  of  j%. 

In  the  following  examples,  cancellation  may  be  employed  by  the 
aid  of  the  Factor  Table. 

25.  What  is  the  value  of  |f  |  x  i|||  X  ||||  ?      Ans.  j%\. 

26.  AVhat  is  the  value  of  ||o  i  x  ||f  i  X  iUi  ?      Ans.  ^. 

27.  What  is  the  value  of  |||f  x  ||f|  X  f  f  |f  ? 

J  ^  a      13  0  3  1 

28.  AYhat  will  7  cords  of  wood  cost  at  S3|  per  cord  ? 

^ns.  $25|. 

29.  What  is  the  value  of  (|y  X  ^f  X  (if?       Ans.   ^tjI^^. 

30.  If  1  horse  eat  |  of  a  bushel  of  oats  in  a  day,  how  many 
bushels  will  10  horses  eat  in  6  days  ?  Ans.  25-|. 

31.  What  is  the  cube  of  12|  ? 

32.  At  S9|  per  ton,  what  will  be  the  cost  of  -|  of  |  of  a  ton  of 
hay?  Ans.  $4. 

33.  At  $'^Q  a  bushel,  what  will  be  the  cost  of  1|  bushels  of 
corn  ? 

34.  When    peaches  are  worth  $|  per  basket,  what  is  ^  of  a 
basket  worth  ? 

35.  A  man  owning  |  of  156|  acres  of  land,  sold  -^  of  |  of  his 
share;  how  many  acres  did  he  sell?  Ans.  47. 

36.  What  is  the  product  of  (f)'  x  Q)'  X  (/^)'  X  (3|y  ? 

Ans.  aT4. 

37.  If  a  family  consume  1|  barrels  of  flour  a  month,  how  many 
barrels  will  6  families  consume  in  8j^^  months? 

38.  What  is  the  product  of  150^— (yQfT21|4-j  of  48|j-— 75, 
multiplied  by  3  x  C|  of  li  x  4)  —  2??  Ans.  342/^^. 


DIVISION.  207 

89.  A  man  at  liis  death  left  liis  wife  §12,500,  wliicli  was  i  of 
'I   of  his   estate;  she   at  her   death   left    |   of  her  share   to   her 
daughter;  what  part  of  the  father's  estate  did  the  daughter  re- 


ceive : 


Jin.,  g^. 


40.  A  owned  |  of  a  cotton  factory,  and  sold  |  of  his  share  to 
B,  who  sold  i  of  v/hat  he  bought  to  C,  who  sold  |  of  what  he 
bought  to  D ;  what  part  of  the  whole  factory  did  each  then  own  ? 

^ns.  A,  i,  ;  B,  J^, ;  C,  ^  _^  A^ 

41.  What  is  the  value  of  2i  x  Hf  ^^  H  X  (|)'+(^3?— ('^3)' • 


A71S. 


^.469 

-'54  0' 


DIVISION. 


194.    1.  Divide  f^  by  3. 

FIRST    OPERATION. 
SECOND    OPERATION. 


21 
^5 


8   =   21 
*^^  —   75 


7 
23 


2.  Dinde  15  by  |. 


Analysis.  In  the  first  ope- 
ration we  divide  the  fraction  by 
3  by  dividing  its  numerator  by 
3,  and  in  the  second  operation 
we  divide  the  fraction  by  3  by 
multiplying  its  denominator  by 
3,  (190,  II  or  III). 


Analysis.  To  divide  by  ?,  we 
must  divide  by  3  and  multiply 
by  7,  (191,  II  or  III). 

In  the  first  operation,  we  first 
divide  15  by  3,  and  then  mul- 
tiply the  quotient  bj  7. 
In  the  second  operation  we  first  multiply  15  by  7,  and  then  divide 
the  product  by  3. 

3.  Divide  X  by  h 

1  o      J    o 


FIRST    OPERATION. 

15  -^  f  =  5  X  7=^35 

SECOND    OPERATION. 

15 -f- 1  =  105 -f- 3  =  35 


FIRST    OPERATION. 
1st  step, 
2d   step,    -^ 


A  -^  3  =  -4, 


4     y    5  —   2  0  4 


SECOND    OPERATION. 
_4      v5   20  4 

^  —  4b  —  y 


Analysis.  To  divide  by 
f ,  we  must  divide  by  3  and 
multiply  by  5,  (191, '  II  or 
III).  In  the  first  operation 
we  first  divide  j-  by  3  by 
multiplying  the  denomina- 
tor by  3.     We  then  multi- 


108  FRACTIONS. 

THIRD  OPERATION.  plj  t^G  result,  $^3,  bj  5,  by 

4  k  multiplying  the  numerator 

£^  X  3  =  I  by  5,  giving  |«  =  |  for  the 

Q  required  quotient.      By  in- 

specting this  operation,  we 
observe  that  the  result,  |§,  is  obtained  by  multiplying  the  denomi- 
nator of  the  given  dividend  by  the  numerator  of  the  divisor,  and  the 
numerator  of  the  dividend  by  the  denominator  of  the  divisor.  Hence, 
in  the  second  operation,  we  invert  the  terms  of  the  divisor,  |,  and 
then  multiply  the  upper  terms  together  for  a  numerator,  and  the 
lower  terms  together  for  a  denominator,  and  obtain  the  same  result  as 
in  the  first  operation.  In  the  third  operation,  we  shorten  the  pro- 
cess by  cancellation. 

We  have  learned  (107)  that  the  reciprocal  of  a  number  is  1 
divided  by  the  number.  If  wg  divide  1  by  |,  we  shall  have  1  — 
I  ==  1  X  I  =  |.     Hence 

195.   The  Reciprocal  of  a  Fraction  is  the  fraction  inverted. 

From  these  principles  and  illustrations  we  derive  the  following 
general 

KuLE.  I.  Reduce  integers  and  mixed  numbers  to  improper 
fractions. 

II.   Midtiply  the  dividend  hy  the  reciprocal  of  the  divisor. 

Notes.  —  1.  If  the  vertical  line  be  U8ed,  the  numerators  of  the  dividend  and 
the  denominators  of  the  divisor  must  be  written  on  the  right  of  the  vertical. 

2.  Since  a  compound  fraction  is  an  indicated  product  of  several  fractions,  its 
reciprocal  may  be  obtained  by  inverting  each  factor  of  the  compound  fraction. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  if  by  4.  f I  X  ^  =  3^^,  Ans. 

2.  Divide  jf  by  5,  and  iff  by  80. 

3.  Divide  10  by  f .  Ans,  35. 

4.  Divide  28  by  |,  and  3  by  /^. 

5.  Divide  56  by  If.  Ans.  36. 

6.  Divide  ^|  by  |. 

7.  Divide  if  by  f ,  4|  by  ^^,,  and  3|  by  5f 

8.  Divide  1|  by  1|.  Ans.  If. 

9.  Divide  l^f  by  |f .  Ans.  If. 


DIVISION.  109 

10.  Divide  f  of  f  by  ?  of  /^. 

OPERATION.  Analysis.    The  dividend, 

3^5-3!  reduced  to  a  simple  fraction, 

7  V    5    5  is  ^ ;    the   divisor,  reduced 

-g  X  1^  —  75  .     f                      .  ' 

1  X   '^  =  ^=11  Ans.  ^^  ^^^  manner,  is  j\  ;  and 


Or, 


^  divided  by  j%  is  14,  the 
quotient  required.     Or,  we 


^  X  -  X   ^  X   ^^  =  11 

5        y        "^         ^  ^  may  apply  the  general  rule 

directly  by  inverting  both  factors  of  the  divisor. 

Note  3.  —  The  second  method  of  solution  given  above  has  two  advantilgea. 
1st,  It  gives  the  answer  by  a  single  operation;  2d,  It  afifords  greater  facility  for 
cancellation. 

11.  Divide  4  of  /^  by  j\  of  /g.  Ans.   1. 

12.  Divide  /^  of  f^  by  |  of  .\.  Ans.  Igi 

13.  Divide  2^  x  7i  by  8i  x  8-^%. 

14.  Divide  11  by  |  X  5|  x  7. 

15.  Divide  31  x  19  by  1  X  7|  x  If.  -Ans.  25. 

16.  Divide  y'V  X  If  by  i  X  I  X  3%  X  |f  X  f  1- 

An,^.  3|f. 

17.  Divide  11^  by  f^^^.  Ans.  1^. 

18.  Divide  3,\Wr  by  U  X  ||  X  g|.  Ans,  ^|. 

19.  Divide  i  X  I  X  I  X  -I  by  I  X  -?  X  i  X  I  X  T%. 

51 

20.  What  is  the  value  of  -f  ? 

OPERATION. 
^  _!.  U     V    JL  5   11 

Analysis.     The  fractional  form  indicates  division,  the  numerator 

being  the  dividend  and  the  denominator  the  divisor,  (168,  II) ;  hence, 

we  reduce  the  mixed  numbers  to  improper  fractions,  and  then  treat 

the  denominator,  2/,  as  a  divisor,  and  obtain  the  result,  IJ,  by  the 

general  rule  for  division  of  fractions. 

51  y 

Note  4. — Expressions  like  —  and  —  are  sometimes  cnWed  complex  fractiona. 

5.  In  the  reduction  of  complex  fractions  to  simple  fractions,  if  either  the 
numerator  or  denominator  consists  of  one  or  more  parts  connected  by  -f  or  — , 
the  operations  indicated  by  these  signs  must  first  be  performed,  and  afterward 
the  division. 

21.  What  is  the  value  of  f  ?  Ans.  ^. 

10 


110  FRACTIONS. 

22.  What  is  the  value  of  -f    ^  l\?  Ans.  2. 

28.   What  is  the  value  of  -^-^t_M  ?  Ans.  7^. 


3  . 


1    5 


24.  Eeduce  ^ ^  to  its  simplest  form. 


'-  + 


25.  Reduce  ^ ^|  to  its  simplest  form. 

3    >^    7 

,5    Qf    3 

26.  Eeduce  7^,^ — p--  to  its  simplest  form.  Ans.  ||. 

27.  If  7  pounds  of  coffee  cost  $|,  how  much  will  1  pound  cost? 

28.  If  a  boy  earn  $|  a  day,  how  many  days  will  it  take  him  to 
earn  $(jl  ?  Ans.   17f 

29.  If  I  of  an  acre  of  land  sell  for  $30,  what  will  an  acre  sell 
for  at  the  same  rate  ?  Ans.   $67^. 

30.  At  ^  of  I  of  a  dollar  a  pint,  how  much  wine  can  be  bought 
for  $^%  ?  ^  Ans.   2|  pints. 

31.  If  -^2^  of  a  barrel  of  flour  be  worth  $21,  how  much  is  1 
barrel  worth  ?  Ans.   $7|. 

32.  Bought  I  of  q  cords  of  wood,  for  |  of  ^   of  $30;  what 
was  1  cord  worth  at  the  same  rate?  Ans.  $4y^. 

33.  If  235^  acres  of  land  cost  $1725|,  how  much  will  125^ 
acres  cost?  Ans.   $918|lf. 

34.  Of  what  number  is  26^  the  |  part?  Ans.   31^. 

35.  The  product  of  two  numbers  is  27,  and  one  of  them  is  2| ; 
what  is  the  other  ? 

36.  By  what  number  must  you  multiply  16}^  to  produce  148|  ? 

37.  What  number  is  that  which,  if  multiplied  by  |  of  g  of  2, 
will  produce  I  ?  Ans.   l|i. 

88.  Divide  720  -  (§  X  28^^71)  by  40|  +  (/,  -  f )  x  Q)*. 

39.  What  is  the  value  of  (b^  X  (ff  +  |  of  jV  ^  (l7^  —  | 

+  FKf)'x5)? 

40    Divide    i°^(t)'x3j  |of5|^  ,,,„, 


GREATEST  COMMON  DIVISOR.  m 


GREATEST  COMMON  DIVISOR  OF  FRACTIONS. 
196.   The  Greatest  Common  Divisor  of  two  or  more  fractions 
is  the  greatest  number  which  will  exactly  divide  each  of  them, 
giving  a  whole  number  for  a  quotient. 

NoTK.  —  The  definition  of  .in  exact  divisor,  (128),  is  general,  and  applies  to 
fractions  as  well  as  to  integers. 

lOT.  In  the  division  of  one  fraction  by  another  the  quotient 
will  be  a  whole  number,  if,  when  the  divisor  is  inverted,  the  two 
lower  terms  may  both  be  canceled.  This  will  be  the  case  w^hen 
the  numerator  of  the  divisor  is  exactly  contained  in  the  numerator 
of  the  dividend,  and  the  denominator  of  the  divisor  exactly 
contains,  or  is  a  multiple  ofy  the  denominator  of  the  dividend. 
Hence, 

I.  A  fraction  is  an  exact  divisor  of  a  given  fraction  w^hen  its 
numerator  is  a  divisor  of  the  given  numerator ,  and  its  denominator 
is  a  viultiple  of  the  given  denominator.     And, 

II.  A  fraction  is  a  common  divisor  of  two  or  more  given  frac- 
tions when  its  numerator  is  a  common  divisor  of  the  given  nume- 
rators^ and  its  denominator  is  a  common  midtij^le  of  the  given 
denominators.     Therefore, 

III.  The  greatest  common  divisor  of  two  or  more  given  frac- 
tions is  a  fraction  whose  numerator  is  the  greatest  common  divisor 
of  the  given  numerators^  and  whose  denominator  is  the  least  com- 
mon multiple  of  the  given  denominators. 

1.  What  is  the  greatest  common  divisor  of  |,  -^r^,  and  j|? 

Analysis.     The  greatest  common  divisor  of  5,  5,  and  15,  the  given 
numerators,  is  5.     The  least  common  multiple  of  6,  12,  and  16,  the 
given  denominators,  is  48.     Therefore  the  greatest  common  divisor 
of  the  given  fractions  is  5=^,  Ans.     (III). 
Proof. 

■jAj  -^-  ^^^  =  4  >  Prime  to  each  other. 

T?    •    45  —  "'  -^ 
1^8.    From  these  principles  and  illustrations,  we  derive  the 
following 


112  FRACTIONS. 

Rule.  Find  the  greatest  common  divisor  of  the  given  nume- 
rators for  a  new  numerator,  and  the  least  common  midtiple  of  the 
given  denominators  for  a  neiv  denominator.  This  fraction  will  he 
the  greatest  common  divisor  sought. 

Note. — Whole  and  mixed  numbers  must  first  be  reduced  to  improper  fractions, 
and  all  fractions  to  their  lowest  terms. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  ^,  i|,  and  ^f  ? 

Ans.  -^1^. 

2.  What  is  the  greatest  common  divisor  of  31,  1^,  and  |^? 

3.  What  is  the  greatest  common  divisor  of  4^  2|,  2|,  and  -^j^  ? 

Ans.  ,2_. 

4.  What  is  the  greatest  common  divisor  of  1091  and  122|  ? 

5.  What  is  the  length  of  the  longest  measure  that  can  be 
exactly  contained  in  each  of  the  two  distances,  18|  feet  and  57^ 
feet?  Ans.  2-^^  feet. 

6.  A  merchant  has  three  kinds  of  wine,  of  the  first  134f  gal- 
lons, of  the  second  128|  gallons,  of  the  third  1151  gallons;  he 
wishes  to  ship  the  same  in  full  casks  of  equal  size;  what  is 
the  least  number  he  can  use  without  mixing  the  different  kinds 
of  wine  ?     How  many  kegs  will  be  required  ?  Ans.  59. 

LEAST  COMMON  MULTIPLE  OF  FRACTIONS. 

199.  The  Least  Common  Multiple  of  two  or  more  fractions  is 
the  least  number  which  can  be  exactly  divided  by  each  of  them, 
giving  a  whole  number  for  a  quotient. 

900.  Since  in  performing  operations  in  division  of  fractions 
the  divisor  is  inverted,  it  is  evident  that  one  fraction  will  exactly 
contain  another  when  the  numerator  of  the  dividend  exactly  con- 
tains the  numerator  of  the  divisor,  and  the  denominator  of  the 
dividend  is  exactly  contained  in  the  denominator  of  the  divisor 
Hence, 

I.  A  fraction  is  a  multiple  of  a  given  fraction  when  its  nume- 
rator is  a  multiple  of  the  given  numerator,  and  its  denominator  is 
a  divisor  of  the  given  denominator.     And 


LEAST  COMMON  MULTIPLE.  113 

II.  A  fraction  is  a  common  multiple  of  two  or  more  given  frac- 
tions when  its  numerator  is  a  common  multiple  of  the  given  nume- 
ratorSy  and  its  denominator  is  a  common  divisor  of  the  given 
denominators.     Therefore, 

III.  The  least  common  multiple  of  two  or  more  given  fractions 

is  a  fraction  whose  numerator  is  the  least  common  multiple  of  the 

given  numerators j  and  whose  denominator  is  the  greatest  common 

divisor  of  the  given  denominators. 

NoTK. — The  least  whole  number  that  will  exactly  contain  two  or  more  given 
fractions  in  their  lowest  terms,  is  the  least  common  multiple  of  their  numera- 
tors, (193,  Note  2). 

1.  What  is  the  least  common  multiple  of  |,  -f^j  and  ||? 

Analysis.  The  least  common  multiple  of  3,  5,  and  15,  the  given 
numerators,  is  15 ;  the  greatest  common  divisor  of  4,  12,  and  16,  the 
given  denominators,  is  4.  Hence,  the  least  common  multiple  of  the 
given  fractions  is  '/  =  3 J,  Ans.  (III). 

301.  From  these  principles  and  illustrations  we  derive  the 
following 

KuLE.  Find  the  least  common  multiple  of  the  given  numerators 
for  a  new  numerator y  and  the  greatest  common  divisor  of  the  given 
denominators  for  a  new  denominator.  This  fraction  will  he  the 
least  commo7i  midtiple  sought. 

Note. — Mixed  numbers  and  integers  should  be  reduced  to  improper  fractions, 
and  all  fractions  to  their  lowest  terms,  before  applying  the  rule. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  least  common  multiple  of  |,  y^^,  ||,  and  3^^^ . 

Ans.  11^. 

2.  What  is  the  least  common  multiple  of  ^^,  ||,  and  ||  ? 

3.  What  is  the  least  common  multiple  of  2||,  1|X,  and  y^^^^  ? 
%    4.  What  is  the  least  common  multiple  of  1,  |,  |,  i,  |,  |,  |,  |, 

and  -^%  ?  ^  Ans.  2520. 

5.  The  driving  wheels  of  a  locomotive  are  ISy^^j  feet  in  circum- 
ference, and  the  trucks  9|  feet  in  circumference.  What  distance 
must  the  train  move,  in  order  to  bring  the  wheel  and  truck  in  the 
same  relative  positions  as  at  starting  ?  Ans.  459|  feet. 

10 '^  H 


114  FRACTIONS. 


TROMISCUOUS    EXAMPLES. 

1.  Change  J  of  |  to  an  equivalent  fraction  having  135  for  its 
denominator.  Ans.  ■^^■^. 

2.  Ptcduue  |,  1, 1^  and  j-l  to  equivalent  fractions^  whose  denom- 
inators shall  be  48. 

3.  Find  the  least  common  denominator  of  1^,  |,  2,  ^-q,  |  of  |, 

4  nf  1 
g  01   4. 

p  of  ^  §-  of  ^    . 

4.  The  sum  of  ^— -^  and  .;'    ,.  ,^.  is  equal  to  how  many  times 

their  difference  ?  Ans.  2. 

543  |5 

5.  The  less  of  two  numbers  is  - — tPk-^j  and  their  difference  -~  ; 

5  o^  H  Tg 

what  is  the  greater  number  ?  Ans.  34y^g3^ 

6.  "What  number  multiplied  by  |  of  |  X  3|j  w^il  produce  ||  ? 

^/iS.    |. 

7.  Find  the  value  of  ^-^  x  ^'  +  (l^  +  A)  -f-  (3  +  4) 

8.  AVhat  number  diminished  by  the  diiiercnce  between  ^  and  ^ 
of  itself,  leaves  a  remainder  of  144  ?  ^7^5.  283 1. 

9.  A  person  spending  -J,  |,  and  ^  of  his  money,  had  §119  left; 
how  much  had  he  at  first? 

10.  What  will  1  of  10|  cords  of  wood  cost,  at  ^^^  of  $42  per 
cord?  ^  ^Ans.  §31^. 

11.  There  are  two  numbers  whose  difference  is  25j'C,  and  one 

1 

Ans.  63|  and  89f^. 

12.  Divide  $2000  between  two  persons,  so  that  one  shall  have 
^  as  much  as  the  other.  Ans.  §1125  and  §875. 

13.  If  a  man  travel  4  miles  in  |  of  an  hour,  how  far  would  he 
travel  in  1^  hours  at  the  same  rate  ?  Ans.   10  miles. 

14.  At  §5  a  yard,  how  many  yards  of  silk  can  be  bought  for 
§1G|? 

15.  How  many  bushels  of  oats  worth  $|  a  bushel,  will  pay  for 
I  of  a  barrel  of  flour  at  §7|  a  barrel  ? 


PROMISCUOUS  EXAMPLES. 


115 


16.  If  f  of  a  bushel  of  barley  be  worth  |  of  a  bushel  of  corn, 
and  corn  be  worth  $'j  per  bushel,  how  many  bushels  of  barley  will 
$15  buy?  '  Ans.  18. 

17.  If  48  is  I  of  some  number,  what  is  |  of  the  same  number? 

18.  If  cloth  1|  yards  in  breadth  require  20i  yards  in  length  to 
make  a  certain  number  of  garments,  how  many  yards  in  length 
will  cloth  I  of  a  yard  wide  require  to  make  the  same  ? 

19.  A  gentleman  owning  |  of  an  iron  foundery,  sold  i  of  his 
share  for  ?2570| ;  how  much  was  the  whole  foundery  worth  ? 

A?u.  $51411. 

20.  Suppose  the  cargo  of  a  vessel  to  be  worth  $10,000,  and  | 
of  I  of  -j^0  of  the  vessel  be  worth  i  of  |  of  If  of  the  cargo;  what 
is  the  whole  value  of  the  ship  and  cargo  ?  Ans.  $22000, 

21.  A  gentleman  divided  his  estate  among  his  three  sons  as  fol- 
lows :  to  the  first  he  gave  |  of  it;  to  the  second  |  of  the  remain- 
der. The  difference  between  the  portions  of  first  and  second  was 
$500.  What  was  the  whole  estate,  and  how  much  was  the  third 
son's  share  ?  .         f  Whole  estate,  $12000. 

I  Third  son's  share,  $2500. 

22.  If  7^  tons  of  hay  cost  $60,  how  many  tons  can  be  bought 
for  $78,  at  the  same  rate  ? 

23.  If  a  person  agree  to  do  a  job  of  work  in  30  days,  what  part 
of  it  ought  he  to  do  in  16^  days?  Ans.  ^^. 

24.  A  father  divided  a  piece  of  land  among  his  three  sons ;  to 
the  first  he  gave  12 J;  acres,  to  the  second  |  of  the  whole,  and  to 
the  third  as  much  as  to  the  other  two;  how  many  acres  did  the 
third  have  ?  Ans.  49  acres. 

25.  If  I  of  6  bushels  of  wheat  cost  $4^,  how  much  will  f  of  1 
bushel  cost  ? 

^  26.  A  man  engaging  in  trade  lost  |  of  his  money  invested,  after 

which  he  gained  $740,  when  he  had  $3500;  hovf  much  did  he 

lose?  '  Ans.  $1840. 

27.  A  cistern  being  full  of  water  sprung  a  leak,  and  before  it 

could  be  stopped,  |  of  the  water  ran  out,  but  |  as  much  ran  in  at 

the  same  time ;  what  part  of  the  cistern  was  emptied  ? 

Ans.  |. 


116  PEACTIONS. 

28.  A  can  do  a  certain  piece  of  work  in  8  days,  and  B  can  do 
the  same  in  6  days ;  in  what  time  can  both  together  do  it  ? 

Ans.  8|  days. 

29.  A  merchant  sold  5  barrels  of  flour  for  $32 J,  which  was  | 
as  much  as  he  received  for  all  he  had  left,  at  §4  a  barrel  •  how 
many  barrels  in  all  did  he  sell  ?  ji^s.  18. 

30.  What  is  the  least  number  of  gallons  of  wine,  expressed  by 
a  whole  number,  that  will  exactly  fill,  without  waste,  bottles  con- 
taining either  |,  |,  |,  or  |  gallons  ?  Ans.   60. 

31.  A,  B,  and  C  start  at  the  same  point  in  the  circumference 
of  a  circular  island,  and  travel  round  it  in  the  same  direction.  A 
makes  |  of  a  revolution  in  a  day,  B  j\,  and  C  /j.  In  how  many 
days  will  they  all  be  together  at  the  point  of  starting  ? 

Ans.  178}  days. 
82.  Two  men  are  64|  miles  apart,  and  travel  toward  each  other; 
when  they  meet  one  has  traveled  5}  miles  more  than  the  other; 
how  far  has  each  traveled  ? 

A72S.  One  29 f  miles,  the  other  35 J  miles. 

33.  There  are  two  numbers  whose  sum  is  ly^^,  and  whose  dif- 
ference is  I;  what  are  the  numbers?  Ajis.  |  and  r^-^, 

34.  A,  B,  and  C  own  a  ferry  boat;  A  owns  ^y^  of  the  boat, 
and  B  owns  -^^^  of  the  boat  more  than  C.  What  shares  do  B  and 
C  own  respectively?  Ans.  B,  -f^;  C,  ^^. 

35.  A  schoolboy  being  asked  how  many  dollars  he  had,  replied, 
that  if  his  money  be  multiplied  by  ||,  and  ^  of  a  dollar  be  added 
to  the  product,  and  |  of  a  dollar  taken  from  the  sum,  this  remainder 
divided  by  3^^^  would  be  equal  to  the  reciprocal  of  |  of  a  dollar. 
How  much  money  had  he  ? 

36.  If  a  certain  number  be  increased  by  If,  this  sum  diminished 
by  |,  this  remainder  multiplied  by  5|,  and  this  product  divided  b|k 
1 1,  the  quotient  will  be  7}  ;  what  is  the  number?  Ans.  ^^. 

37.  If  I  of  4  of  3}  times  any  number  be  multiplied  by  |,  the 
product  divided  by  |,  the  quotient  increased  by  4  J,  and  the  sum 
diminished  by  |  of  itself,  the  remainder  will  be  how  many  times 
the  number  ?  Ans,  6 j\^^  times. 


r 


NOTATION  AND  NUMERATION.  jj^ 

DECIMAL  FRACTIONS. 
NOTATION  AND  NUMERATION. 


303.  A  Decimal  Fraction  is  one  or  more  of  the  decimal 
divisions  of  a  unit. 

NoTRS. — 1.  The  word  decimal  is  derived  from  the  Latin  decern,  which  signi- 
fies fen. 

2.  Decimal  fractions  are  commonly  called  decimals, 

^03.  In  the  formation  of  decimals,  a  simple  unit  is  divided 
into  ten  equal  parts,  forming  decimal  units  of  the  first  order,  or 
tenths,  each  tenth  is  divided  into  ten  equal  parts,  forming  decimal 
units  of  the  second  order,  or  hundredths;  and  so  on,  according  to 
the  following 

TABLE  OF  DECIMAL  UNITS. 

1  single  unit  equals  10  tenths ; 

1  tenth  *'  10  hundredths ; 

1  hundredth         "  10  thousandths; 

1  thousandth        "  10  ten  thousandths^ 
etc.  etc. 

304.  In  the  notation  of  decimals  it  is  not  necessary  to  employ 
denominators  as  in  common  fractions;  for,  since  the  dificrent 
orders  of  units  are  formed  upon  the  decimal  scale,  the  same  law 
of  local  value  as  governs  the  notation  of  simple  integral  numbers, 
(o"^),  enables  us  to  indicate  the  relations  of  decimals  by  place 
or  position. 

S05j    The  Decimal  Sign  (.)  is  always  placed  before  decimal 
^gures  to  distinguish  them  frcfm  integers.     It  is  commonly  called 
^1^  decimal  point.     When  placed  between  integers  and  decimals 
in  the  same  number,  is  sometimes  called  the  separatrix. 

SOO.  The  law  of  local  value,  extended  to  decimal  units,  as- 
signs the  first  place  at  the  right  of  the  decimal  sign  to  tenths ; 
the  second,  to  hundredths;  the  third^  to  thousandths;  and  so  on, 
as  shown  in  the  followinsc 


118  DECIMALS. 


DECIMAL    NUMERATION    TABLE. 

OQ 
-M 

^  ^  ^  g 


g 

OQ 
O 

DQ 

13 

1 

03 

no 

1 

o 

€4-1 

0 

02 

0 

J3 

0 

c   §   2   i   §  ^ 

.  ".    .  '~,       q^pJ-HiJ".      "-i^,^,^      O^,-^      C^    -^    13 


5732754.57325 


ooooggS    Sggooo 


2©7«  The  denominator  of  a  decimal  fraction,  when  expressed, 
is  necessarily  10,  100,  1000,  or  some  power  of  10.  By  examining 
the  table  it  will  be  seen,  that  the  number  of  places  in  a  decimal  is 
equal  to  the  number  of  ciphers  required  to  express  its  denomi- 
nator. Thus,  tenths  occupy  the  first  place  at  the  right  of  units, 
and  the  denominator  of  j'^  has  one  cipher;  hundredths  in  the 
table  extend  two  places  from  units,  and  the  denominator  of  j^-q 
has  two  ciphers ;  and  so  on. 

S08.  A  decimal  is  usually  read  as  expressing  a  certain  number 
of  decimal  units  of  the  lowest  order  contained  in  the  decimal. 
Thus,  5  tenths  and  4  hundredths,  or  .54,  may  be  read,  fifty-four 
hundredths.     For,  j\  +  j^^  =  -f-^%. 

20^.  From  the  foregoing  explanations  and  illustrations  we 
derive  the  following  ^^ 

PRINCIPLES    OF    DECIMAL    NOTATION    AND    NUMERATION, 

I.  Decimals  are  governed  by  the  same  law  of  local  value  that 
governs  the  notation  of  integers. 

II.  The  different  orders  of  decimal  units  decrease  from  left  to 
right,  and  increase  from  right  to  left,  in  a  tenfold  ratio. 


NOTATION  AND  NUMERATION.  119 

III.  The  value  of  any  decimal  figure  depends  upon  the  place 
it  occupies  at  the  right  of  the  decimal  sign. 

,  TV.  Prefixing  a  cipher  to  a  decimal  diminishes  its  value  ten- 
fold, since  it  removes  every  decimal  figure  one  place  to  the  right. 

y.  Annexing  a  cipher  to  a  decimal  does  not  alter  its  value, 
since  it  does  not  change  the  place  of  any  figure  in  the  decimal. 

YI.  The  denominator  of  a  decimal,  when  expressed,  is  the 
unit,  1,  with  as  many  ciphers  annexed  as  there  are  places  in  the 
decimal. 

YII.  To  read  a  decimal  requires  two  numerations ;  first,  from 
units,  to  find  the  name  of  the  denominator;  second,  towards  units, 
to  find  the  value  of  the  numerator. 

SIO.  Having  analyzed  all  the  principles  upon  which  the 
writing  and  reading  of  decimals  depend,  we  will  now  present  these 
principles  in  the  form  of  rules. 

RULE    FOR    DECIMAL    NOTATION. 

I.  Write  the  decimal  the  same  as  a  whole  number ^  placing 
cipher's  in  the  place  of  vacant  orders^  to  give  each  significant  figure 
its  true  local  value. 

II.  Place  the  decimal  point  he/ore  the  first  figure. 

RULE    FOR    DECIMAL    NUMERATION. 
T.  Numerate  from  the  decimal  pointy  to  determine  the  denomi- 
nator. 

II.  Numerate  towards  the  decimal  point,  to  determine  the  nu- 
merator. 

III.  Read  the  decimal  as  a  whole  number,  giving  it  the  name 
of  its  lowest  decimal  unit,  or  right  hand  figure. 

wL  EXAMPLES    FOR   PRACTICE. 

Express  the  following  decimals  by  figuiHes,  according  to  the 
decimal  notation. 

1.  Five  tenths.  Ans.  .5. 

2.  Thirty-six  hundredths.  Ans.  .36, 

3.  Seventy-five  ten-thousandths.  Ans.  .0075. 


120 


DECIMALS. 


4.  Four  hundred  ninety-six  thousandths. 

5.  Three  hundred  twenty-five  ten-thousandths. 

6.  One  millionth. 

7.  Seventy-four  ten-million ths. 

8.  Four  hundred  thirty-seven  thousand  five  hundred  forty- 
nine  millionths. 

9.  Three  million  forty  thousand  ten  ten-million  ths. 

10.  Twenty-four  hundred-millionths. 

11.  Eight  thousand  six  hundred  forty-five  hundred-thousandths. 

12.  Four  hundred  ninety-five  million  seven  hundred  five  thou- 
sand forty-eight  billionths. 

13.  Ninety-nine  thousand  nine  ten-billionths. 

14.  Four  million  seven  hundred  thirty-five  thousand  nine  hun- 
dred one  hundred-millionths. 

15.  One  trillionth. 

16.  One  trillion  one  billion  one  million  one  thousand  one  ten- 
trillionths. 

17.  Eight  hundred  forty-one  million  five  hundred  sixty-three 
thousand  four  hundred  thirty-six  trillion  ths. 

18.  Nine  quintillionths. 

Express  the  following  fractions  and  mixed  numbers  decimally : 

46.4. 


19.  j%.                   Ans.  .3. 

25. 

46/^.          Ans.  4 

20.  I'AV 

99            85 
^-'-    TCJOOUI)- 
9Q        100004 
^^-    TTJOO^UIJ- 

26. 
27. 

28. 
29. 
30. 

205,-. 

'^^TUO(T^%(jnTJO- 
^"t^55T5T505TJTJTJT5TJ 

Read  the  following  numbers : 

31.  .24. 

38. 

8.25. 

32.  .075. 

39. 

75.368. 

33.  .503.            ^ 

40. 

42.0637. 

34.  .00725. 

41. 

8.0074. 

35.  .40000004. 

42. 

30.4075. 

36.  .0000256. 

43. 

26.00005. 

37.  .0010075, 

44. 

100.00000001. 

REDUCTION.  121 

REDUCTION. 
CASE  I. 

311.  To  reduce  decimals  to  a  common  denominator. 
1.  Reduce  .5,  .24,  .7836  and  .375  to  a  common  denominator. 

OPER\Tiox  Analysis.     A  common  denominator  must  contain 

F.()r\f)  as  many  decimal  places  as  are  equal  to  the' greatest 

oiQQ  number  of  decimal  figures  in  any  of  the  given  deci- 

.7836  mals.     We  find  that  the  third  number  contains  four 

.3750  decimal  places,  and  hence  10000  must  be  a  common 

denominator.     As  annexing  ciphers  to  decimals  does 

not  alter  their  value,  we  give  to  each  number  four  decimal  places,  by 

annexing  ciphers,  and  thus  reduce  the  given  decimals  to  a  common 

denominator.     Hence, 

Rule.  Give  to  each  number  the  same  number  of  decimal 
placeSj  by  annexing  ciphers. 

Notes. — 1.  If  the  numbers  be  reduced  to  the  denominator  of  that  one  of  the 
given  numbers  having  the  greatest  number  of  decimal  places,  they  will  have 
their  least  common  decimal  denominator. 

2.  An  integer  m;iy  readily  be  reduced  to  decimals  by  placing  the  decimal 
point  after  units,  and  annexing  ciphers  ;  one  cipher  reducing  it  to  tenths,  two 
ciphers  to  hundredths,  three  ciphers  to  thousandths,  and  so  on. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  .18,'  .456,  .0075,  .000001,  .05,  .3789,  .5943786,  and 
.001  to  their  least  common  denominator. 

2.  Reduce  12  thousandths,  185  millionths,  936  billionths,  and 
7  trillionths  to  their  least  common  denominator. 

3.  Reduce  57.3,  900,  4.7555,  and  100.000001  to  their  least 
common  denominator. 


\ 


CASE   II. 

313.   To  reduce  a  decimal  to  a  common  fraction. 
1.   Reduce  .375  to  an  equivalent  common^^ction. 

^ „,^^,  Analysis.    Writinsr  the  decimal 

OPERATION.  ° 

r>^^-         .^^  «  fiojures,  .375,  over  the  common  de- 

•^  <  ^  —  1 OUU  —  #•  nominator,  1000,  we  have  yVA  =  f . 

Hence, 

11 


122  DECIMALS. 

HuLE.      Omit  the  decimal  point,  and  supply  the  proper  denomi- 
nator. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  .75  to  a  common  fraction.  Ans.  |. 

2.  Reduce  .625  to  a  common  fraction.  Ans.  |. 

3.  Reduce  .12  to  a  common  fraction. 

4.  Reduce  .68  to  a  common  fraction. 

5.  Reduce  .5625  to  a  common  fraction. 

6.  Reduce  .024  to  a  common  fraction.  Ans.  y|^. 

7.  Reduce  .00032  to  a  common  fraction.  Ans.  ^j^-^* 

8.  Reduce  .002624  to  a  common  fraction.  Ans-  y/glj^- 

9.  Reduce  .13|  to  a  common  fraction. 

OPERATION. 

131  —  ll^  —     4  0    ^     2 
.xc»3  ^^^  -g^^  —    jj^. 

Note. — The  decimal  .13^  may  properly  be  called  a  complex  decimal. 

10.  Reduce  .57^  to  a  common  fraction.  Ans.  ^. 

11.  Reduce  .66|  to  a  common  fraction.  Ans.  |. 

12.  Reduce  .444^  to  a  common  fraction. 

13.  Reduce  .024|  to  a  common  fraction.  Ans.  yfj^. 

14.  Reduce  .984|  to  a  common  fraction. 

15.  Express  7.4  by  an  integer  and  a  common  fraction. 

Ans.  7|. 

16.  Express  24.74  by  an  integer  and  a  cojnmon  fraction. 

17.  Reduce  2.1875  to  an  improper  fraction.  Ans.  ||. 

18.  Reduce  1.64  to  an  improper  fraction. 

19.  Reduce  7.496  to  an  improper  fraction.  Ans.  f| 


7 


CASE    III. 

213.   To  reduce  a  common  fraction  to  a  decimal,     ip 
1.  Reduce  |  to  tIs  equivalent  decimal. 

FIRST  OPERATION.  ANALYSIS.     "VYc   first    annex 

5  5  000  62  5  C9^  ^^^  same  number  of  ciphers  to 

H  —  H 0^^  —  TTJ^U  =  -^-^  i^oth  terms  of  the  fraction ;  this 

does  not  alter  its  value,  (174, 


REDUCTION.  123 

SECOND  OPERATION.  Ill) ;  we  then  divide  both  re- 

8  ^  5.000  suiting  terms  by  8,  the  siji;nifi- 

—~~  cant  figure  of  the  denominator, 

*   "^  and  obtain  the  decimal  denom- 

inator, 1000.     Omitting  the  denominator,  and  prefixing  the  sign,  we 
have  the  equivalent  decimal,  .025. 

In  the  second  operation,  we  omit  the  intermediate  steps,  and  obtain 
the  result,  practically,  by  annexing  the  three  ciphers  to  the  nume- 
,rator,  5,  and  dividing  the  result  by  the  denominator,  8. 

2.  Reduce  j|-  to  a  decimal. 

OPERATION.  Analysis.     Dividing  as  in  the  former  ex- 

125  )  8.000  ample,  we  obtain  a  quotient  of  2  figures,  24. 

024  But  since  3  ciphers  have  been  annexed  to  the 

numerator,  3,  there  must  be  three  places  in  the 

required  decimal ;  hence  we  prefix  1  cipher  to  the  quotient  figures, 

24.     The  reason  of  this  is  shown  also  in  the  following  operation. 

3     3000      24     __    02-t 

8141.    From  these  illustrations  we  derive  the  following 
E-ULE.     I.  Aimex  ciphers  to  the  numerator^  and  divide  hy  the 

denominator. 

II.  Point  off  as  many  decimal  places  in  the  result  as  arc  equal 

to  the  number  of  ciphers  annexed. 

Note.-  If  the  division  is  not  exact  when  a  sufficient  number  of  decimal 
figures  have  been  obtained,  the  sign,  +,  may  bo  annexed  to  the  decimal  to  indi- 
cate that  there  is  still  a  remainder.  When  this  remainder  is  such  that  the  next 
decimal  figure  would  be  5  or  greater  than  5,  the  last  figure  of  the  terminated 
decimal  may  be  increased  by  1,  and  the  sign,  — ,  annexed.  And  in  general,  + 
denotes  that  the  written  decimal  is  too  small,  and  —  denotes  that  the  written 
decimal  is  too  large ;  the  error  always  being  less  than  one  half  of  a  unit  in  the 
last  place  of  the  decimal. 

EXAMPLES  FOR  PRACTICE. 
1.  Reduce  |  to  a  decimal.  Ans.  .75. 


2.  Reduce  /^  to  a  decimal.  ^       Ans.  .3125. 

3.  Reduce  Z  to  a  decimal. 


5 

4.  Reduce  ^i  to  a  decimal. 

5.  Reduce  ||  to  a  decimal. 

6.  Reduce  ^^  to  a  decimal.  Ans.   .04. 

7.  Reduce  ^Jj^  to  a  decimal.  Ans,  .068. 


124 


DECIMALS. 


8.  Reduce  ^^  to  a  decimal.  Ans,  .59375 

9.  Reduce  y^g^^^  ^^  ^  decimal.  i 

10.  Reduce  ^^  to  a  decimal.  Arts,  .29167 — . 

11.  Reduce  -^^j^  to  a  decimal. 

12.  Reduce  ||  to  a  decimal.  -4ns.  .767857+. 

13.  Reduce  7^  to  the  decimal  form.  Ans.  7.125. 

14.  Reduce  56/5  to  the  decimal  form.  Ans,  56.078125. 

15.  Reduce  32|  to  the  decimal  form. 

16.  Reduce  .24^  to  a  simple  decimal. 

17.  Reduce  5.78 1§  to  a  simple  decimal. 

18.  Reduce  .3y-^4^  to  a  simple  decimal.  Ans.  .30088. 

19.  Reduce  ^-^^^  to  a  simple  decimal.  Ans.  4.008. 

20.  Reduce  •30y||§^^  to  a  simple  decimal.         '  ' 


AlDDITION. 

313.  Since  the  same  law  of  local  value  extends  both  to  the 
rio'ht  and  left  of  units'  place ;  that  is,  since  decimals  and  simple 
integers  increase  and  decrease  uniformly  by  the  scale  of  ten,  it  is 
evident  that  decimals  may  be  added,  subtracted,  multiplied  and 
divided  in  the  same  manner  as  integers. 

216.     1.  What  is'the  sum  of  4.75,  .246,  37.56  and  12.248  ? 

OPERATION.  Analysis.     We  write  the  numbers  so  that  units  of 

4.75  like  orders,  whether  integral  or  decimal,  shall  stand 

.246  in  the  same  columns ;  that  is,  units  under  units,  tenths 

37.56  under  tenths,  etc.     This  brings  the  decimal  points 

12.248  directly  under  each  other.     Commencing  at  the  right 

54  804  hand,  we  add  each  column  separately,  carrying  1  for 

every  ten,  according  to  the  decimal  scale  ;  and  in  the 
result  we  place  the  decimal  point  between  units  and  tenths,  or  directly 
under  the  decimal  points  in  the  numbers  added.  Hence  the  fol- 
lowing 0^ 

Rule.  I.  Write  the  numbers  so  that  the  decimal  points  shall 
stand  directly  under  each  other, 

II.  Add  as  in  whole  numhersj  and  place  the  decimal  pointy  in 
the  result  J  directly  under  the  points  in  the  numbers  added. 


ADDITION".  125 

EXAMPLES    FOR    PRACTICE. 

1.  Add  .375,  .24,  .536,  .78567,  .4637,  and  .57439. 

Ans.  2.97476. 

2.  Add  5.3756,  85.473,  9.2,  46.37859,  and  45.248377. 

Ans.  191.675567. 

3.  Add  .5,  .37,  .489,  .6372,  .47856,  and  .02524. 

4.  Add  .46|,  .325|,  .16^%,  and  .275/^.       Ans.  1.2296625. 

5.  Add  4.6^,  7.32 3L,  5.3784^,  and  2.64878|. 

6.  Add  4.3785,  2|/5f,  and  12.4872.  Ans.  24.9609  +  . 

7.  What  is  the  sum  of  137  thousandths,  435  thousandths,  836 
thousandths,  937  thousandths,  and  496  thousandths  ? 

Ans.  2.841. 

8.  What  is  the  sum  of  one  hundred  two  ten-thousandths,  thir- 
teen thousand  four  hundred  twenty-six  hundred  thousandths,  five 
hundred  sixty-seven  millionths,  three  millionths,  and  twenty-four 
thousand  seven  hundred-thousandths  ? 

9.  A  farm  has  five  corners;  from  the  first  to  the  second  is  34.72 
rods;  from  the  second  to  the  third,  48.44  rods;  from  the  third  to 
the  fourth,  152.17  rods;  from  the  fourth  to  the  fifth,  95.36  rods; 
and  from  the  fifth  to  the  first,  56.18  rods.  What  is  the  whole 
distance  around  the  farm  ? 

10.  Find  the  sum  of  ||,  -^.fjr,  3^/^,  and  yiy^H  i^  decimals,  correct 

to  the  fourth  place.  Ans.  .6GC9  +  . 

Note. — In  the  reduction  of  each  fraction,  carry  the  decimal  to  at  least  the 
fifth  place^  in  order  to  insure  accuracy  in  the  fourth  place. 

11.  A  man  owns  4  city  lots,  containing  16-,^^  rods,  15^^  rods, 
18^1  rods,  and  ll^''^  rods  of  land,  respectively;  how  many  rods 
in  all  ? 

12.  What  is  the  sum  of  4^^  decimal  units  of  the  first  order,  2| 
of  the  second  order,  9^  of  the  third  order,  and  3^V  of  the  fourth 
order?  aIs.  .486929. 

13.  What  is  the  approximate  sum  of  1  decimal  unit  of  the  first 
order,  J-  of  a  unit  of  the  second  order,  |  of  a  unit  of  the  third 
order,  I  of  a  unit  of  the  fourth  order,  i  of  a  unit  of  the  fifth  order, 
J  of  a  unit  of  the  sixth  order,  and  7^  of  a  unit  of  the  seventh  order  ? 

Ans.  .1053605143—. 
11* 


126 


DECIMALS. 


917. 


SUBTRACTION. 
1.  From  4.156  take  .5783. 


OPERATIOX. 

4.1560 

.5783 


Analysis.  We  write  the  given  numbers  as  in  addi- 
tion, reduce  the  decimals  to  a  common  denominator, 
and  subtract  as  in  integers.  Or,  we  may,  in  practice, 
omit  the  ciphers  necessary  to  reduce  the  decimals  to  a 
common  denominator,  and  merely  conceive  them  to  be 
annexed,  subtracting  as  otherwise.  Hence  the  fol- 
lowing 

3.5777 

lluLE.  I.  Write  (he  'numbers  so  that  the  decimal  points  shall 
stand  directly/  iinder  each  other, 

II.  Subtract  as  in  tchole  numbers,  and  place,  the  decimal  point 
in  the  result  directly  under  the  points  in  the  given  numbers. 


S.0717 
Or, 

4.156 

.5783 


EXAMPLES    FOR   PRACTICE. 


9. 

10. 
11. 
12. 


.9876 
.3598 


(2.) 
48.3676 
23.98 


Minuend, 
Subtrahend, 

Remainder,       .6278  24.3876 

From  37.456  take  24.367. 
From  1.0066  take  .15. 
From  1000  take  .001. 
From  36|  take  22^1. 

4  2  o 


From 
From  7 


56|  take  .55j||. 
take  5/5. 


(3.) 
36.5 
35.875632 

.624368 
.4ns.  13.089. 

Ans.  999.999. 
Ans.  14.27. 

Ans.  1.7708  +  . 


From  |§4  take  ^J f. 


From  one  take  one  trillionth.  Ans.  .999999999999. 

A  speculator  having  57436  acres  of  land,  sold  at  different 
times  536.74  acres,  1756.19  acres,  3678.47  acres,  9572.15  acres, 
7536.59  acres,  and  4785.94  acres;  how  much  land  has  he 
remaining  ? 

13.  Find  the  difference  between  f ||f i  and  ;Jf||f,  correct  to 
the  fifth  decimal  place.  ,  Ans.  4.17298+. 


MULTIPLICATION. 


MULTIPLICATION. 


127 


918.  In  multiplication  of  decimals,  the  location  of  the  decimal 
point  in  the  product  depends  upon  the  following  principles  : 

I.  The  number  of  ciphers  in  the  denominator  of  a  decimal  is 
equal  to  the  number  of  decimal  places,  (200,  YI). 

II.  If  two  decimals,  in  the  fractional  form,  be  multiplied  to- 
gether, the  denominator  of  the  product  must  contain  as  many 
ciphers  as  there  are  decimal  places  in  both  factors.     Therefore, 

III.  The  product  of  two  decimals,  expressed  in  the  decimal 
form,  must  contain  as  many  decimal  places  as  there  are  decimals 
in  both  factors. 

1.  Multiply  .45  by  .7. 

OPERATION.  Analysis.   We  first  multiply 

^c  as   in  w^hole   numbers ;    then, 

^T"  since  the  multiplicand  has  2 

0-,  r  decimal  places  and  the  multi- 

plier 1,  we  point  ofi*  2  +  1  =  3 
PROOF.  decimal  places  in  the  product, 

^4^5^  X  7^5  =  -rV/(J  =  -315  (HI).     The   reason  of  this  is 

further  illustrated  in  the  proof, 
a  method  applicable  to  all  similar  cases. 

910.  Hence  the  following 

Rule.  Mxdtiply  as  in  whole  nvmhers,  and  from  the  right  hand 
of  the  product  point  off  as  many  figures  for  decimals  as  there  are 
decimal  places  in  both  factors. 

Notes.  —  1.  If  there  be  not  as  many  figures  in  the  product  as  there  are  deci- 
mals in  both  factors,  supply  the  deficiency  by  prefixing  ciphers. 

2.  To  multiply  a  decimal  by  10,  100,  1000,  etc.,  remove  the  point  as  many 
places  to  the  right  as  there  are  ciphers  on  the  right  of  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  .75  by  .41.  Ans.  .3075. 

2.  Multiply  .436  by  .24. 

3.  Multiply  5.75  by  .35.  Ans.  2.0125.  , 

4.  Multiply  .756  by  .025.  Ans.  .0189. 

5.  Multiply  3. 784  by  2.475. 


128  DECIMALS. 

6.  Multiply  7.23  by  .0156.  An^.  .112788. 

7.  Multiply  .0075  by  .005.  Ans.  .0000375. 

8.  Multiply  324  by  .324. 

9.  Multiply  75.64  by  .225. 

10.  Multiply  5.728  by  100.  Ans,  572.8. 

11.  Multiply  .36  by  1000. 

12.  Multiply  .000001  by  1000000. 

13.  Multiply  .576  by  100000. 

14.  Multiply  7|  by  5^.  Ans.  42.625. 

15.  Multiply  .63^  by  24. 

16.  Multiply  4/^  by  7^%.  Ans.  31.74. 

17.  Find  the  value  of  3.425  x  1.265  x  64.    Ans.  277.288. 

18.  Find  the  value  of  32  x  .57825  x  .25. 

19.  Find  the  value  of  18.375  x  5.7  X  1.001. 

Ans.  104.8422375. 

20.  If  a  cubic  foot  of  granite  weigh  168.48  pounds,  what 
will  be  the  weight  of  a  granite  block  that  contains  271  cubic 
feet? 

21.  When  a  bushel  of  corn  is  worth  2.8  bushels  of  oats,  how 
many  bushels  of  oats  must  be  given  in  exchange  for  36  bushels 
of  corn  and  48  bushels  of  oats  ?  Ans.  148.8. 


CONTRACTED    MULTIPLICATION. 

SSO.  To  obtain  a  given  number  of  decimal  places  in 
the  product. 

It  is  frequently  the  case  in  multiplication,  that  a  greater  number 
of  decimal  figures  is  obtained  in  the  product,  than  is  necessary 
for  practical  accuracy.  This  may  be  avoided  by  contracting  each 
partial  product  to  the  required  number  of  decimal  places. 

To  investigate  the  principles  of  this  method,  let  us  take  the  two 
decimals  .12345  and  .54321,  and  having  reversed  the  order  of  tho 
digits  in  the  latter,  and  written  it  under  the  former,  multiply  each 
figure  of  the  direct  number  by  the  figure  below  in  the  reversed  num- 
ber, placing  the  products  with  like  orders  of  units  in  the  same  column, 
thus : 


CONTRACTED  MULTIPLICATION.  129 


.12345  direct      = 

.12345 

,54321  reversed  = 

:  12345. 

.00UU25  =  .00005  X  .5 

.000016  =  .0004    X  .04 

.000009  =  .003      X  .003 

.000004  =  .02        X  .0002 

.000001  =  .1          X  .00001. 

In  this  operation  we  perceive  that  all  the  products  are  of  the  same 
order ;  and  this  must  always  be,  whether  the  numbers  used  be  frac- 
tional, integral,  or  mixed.  For,  as  we  proceed  from  right  to  left  in 
the  multiplication,  we  pass  regularly  from  lower  to  higher  orders  in 
the  direct  number,  and  from  higher  to  lower  in  the  reversed  number. 
Hence 

2SI.  If  one  number  be  written  under  another  with  the  order 
of  its  digits  reversed,  and  each  figure  of  the  reversed  number  be 
multiplied  by  the  figure  above  it  in  the  direct  number,  the  prod- 
ucts will  all  be  of  the  same  order  of  units. 

1.  Multiply  4.78567  by  3.25765,  retaining  only  3  decimal 
places  in  the  product. 

OPERATION.  Analysis.    Since  the  product 

.  ^c-n'T  of  any  figure  by  units  is  of  the 

4./8o67  "^    ^        .•!     ^  -. 

^fiy'r^  ^  same  order  as  the  figure  multi- 

plied, (82,  II,)  we  write  3,  the 
units  of  the  multiplier,  under 
5,  the  third  decimal  figure  of 
the  multiplicand,  and  the  lowest 
order  to  be  retained  in  the  pro- 
duct ;  and  the  other  figures  of 
the  multiplier  we  write  in  the 
Inverted  order,  extending  to  the  left.  Then,  since  the  product  of  3 
and  5  is  of  the  third  order,  or  thousandths,  the  products  of  the  other 
corresponding  figures  at  the  left,  2  and  8,  5  and  7,  7  and  4,  etc.,  will 
be  thousandths ;  and  we  therefore  multiply  each  figure  of  the 
multiplier  by  the  figures  above  and  to  the  left  of  it  in  the  multipli- 
cand, carrying  from  the  rejected  figures  of  the  multiplicand,  as  fol- 
lows :  3  times  G  are  18,  and  as  this  is  nearer  2  units  than  one  of  the 
next  higher  order,  we  must  carry  2  to  the  first  contracted  product ;  3 
times  5  are  15,  and  2  to  be  carried  are  17 ;  writing  the  7  under  the 
3,  and  multiplying  the  other  figures  at  the  left  in  the  usual  manner, 

I 


14357  = 

4785 

X 

3  +  2 

957  = 

478 

X 

2  +  1 

239  = 

47 

X 

5  +  4 

33  = 

4 

X 

7+A 

3  = 

0 

X 

6+f 

15.589=1=,  Ans. 


130  DECIMALS. 

we  obtain  14357  for  the  first  partial  product.  Then,  beginning  with 
the  next  figure  of  the  multiplier,  2  times  5  are  10,  which  gives  1  to 
be  carried  to  the  second  partial  product ;  2  times  8  are  16,  and  1  to  be 
carried  are  17  ;  writing  the  7  under  the  first  figure  of  the  former  pro- 
duct, and  multiplying  the  remaining  left-hand  figures  of  the  mul- 
tiplicand, we  obtain  957  for  the  second  partial  product.  Then,  5 
times  8  are  40,  which  gives  4  to  be  carried  to  the  third  partial  pro- 
duct ;  5  times  J  are  35  and  4  are  39  ;  writing  the  9  in  the  first  column 
of  the  products,  and  proceeding  as  in  the  former  steps,  we  obtain  239 
for  the  third  partial  product.  Next,  multiplying  by  7  in  the  same 
manner,  we  obtain  33  for  the  fourth  partial  product.  Lastly,  begin- 
ning 2  places  to  the  right  in  the  multiplicand,  6  times  7  are  42 ;  6 
times  4  are  24,  and  4  are  28,  which  gives  3  to  be  carried  to  the  fifth 
partial  product;  6  times  0  is  0,  and  3  to  be  carried  are  3,  which  we 
write  for  the  last  partial  product.  Adding  the  several  partial  pro- 
ducts, and  pointing  ofi"  3  decimal  places,  we  have  15.589,  the  required 
product. 

22S.  From  these  principles  and  illustrations  we  derive  the 
following 

Rule.  I.  Write  the  mnltij)lier  with  the  order  of  its  figures 
reversed,  and  xcifh  the  units'  place  under  that  figure  of  the  midti- 
plicand  which  is  the  lowest  decimal  to  he  retained  in  the  product. 

II.  Find  the  product  of  each  figure  of  the  multiplier  hy  the 
figures  ahove  and  to  the  left  of  it  in  the  midtiplicandj  increasing 
each  partial  2^roduct  hy  as  many  units  as  would  have  heen  carried 
from  the  rejected  part  of  the  mul^ff^icandy  and  one  more  when  the 
highest  figure  in  the  rejected  jiart  of  any  product  is  b  or  greater 
than  5 ;  and  write  these  partial  p)roducts  with  the  lowest  figure  of 
each  in  the  same  column. 

III.  Add  the  partial  products  J  and  from  the  right  hand  of  the 

resxdt  j^oint  off  the  required  numher  of  decimal  figures. 

NoTKS. — 1.  In  ol-taining  the  number  to  he  carried  to  each  contrncted  partial 
product,  it  is  jrcnerally  necessary  to  multiply  (mentally)  only  one  figure  at  the 
riirht  of  the  figure  above  the  multiplyino^  fl<;^ure;  but  when  the  figures  are  large, 
the  mtiltiplication  should  comnienae  at  least  two  places  to  the  right. 

2.  Observe,  that  when  the  number  of  units  in  the  highest  order  of  the  rejected 
part  of  the  product  is  between  5  and  15,  carry  1;  if  between  15  and  25  carry 
2;  if  between  25  and  35  carry  3;  and  so  on. 

3.  There  is  always  a  liability  to  an  error  of  one  or  two  units  in  the  last  place: 
and  as  the  answer  may  be  either  too  great  or  too  small   by  the  amount  of  thig 


CONTRACTED  MULTIPLICATION.  1^1 

error,  the  uncertainty  may  be  indicated  by  the    double  sign,  ±,  read,  plus,  or 
vnnus,  and  placed  after  the  product. 

4.  When  the  number  of  decimal  places  in  the  multiplicand  is  less  than  the 
number  to  be  retained  in  the  product,  supply  the  deficiency  by  annexing  ciphers. 

EXAMPLES    FOR    PRACTICE. 

1.  Multiply  230.45  by  32.46357,  retaining  2  decimal  places, 
and  2.563789  by  .0347263,  retaining  0  decimal  places  in  the 
product. 

OPERATION".  OPERATION. 

236.450  2.563789 

75364.23  362  7430. 

709350  76914 

47290  10255 

9458  1795 

1419  51 

71  15 

12  1 


2  .089031 


7676.02  ± 

2.  Multiply  36.275  by  4.S678,  retaining  1  decimal  place  in  the 
product.  Ans.  158.4  zh. 

3.  Multiply  .24367  by  36.75,  retaining  2  decimal  places  in  the 
product. 

4.  Multiply  4256.785  by  .00564,  rejecting  all  beyond  the  third 
decimal  place  in  the  product.^  Ans.  24.008  =i=. 

5.  Multiply  357.84327  by  1.007806,  retaining  4  decimal  places 
in  the  product. 

6.  Multiply  400.756  by  1.367583,  retaining  2  decimal  places  in 
the  product.  ^4/^^.   548.07  =i=. 

7.  Multiply  432.5672  by  1.0666666,  retaining  3  decimal  places 
in  the  product. 

8.  Multiply  48.4367  by  2^^^,  extending  the  product  to  three 
decimal  places.  Ans.  103.418  ±. 

9.  Multiply  7jf3  by  3|J§,  extending  the  product  to  three 
decimal  places. 

10.  The  first  satellite  of  Uranus  moves  in  its  orbit  142.8373  + 


182  DECIMALS. 

1 
degrees  in  1  day;  find  how  many  degrees  it  will  move  in  2.52035 

days,  carrying  the  answer  to  two  decimal  places. 

A71S,  860.00  degrees. 

11.  A  gallon  of  distilled  water  weighs  8.33888  pounds ;  how 
many  pounds  in  35.8756  gallons?  Ans.  299.16  db  pounds. 

12.  One  French  metre  is  equal  to  1.09356959  English  yards; 
how  many  yards  in  478.7862  metres.         Ans.  523.58  db  yards. 

13.  The  polar  radius  of  the  earth  is  6356078.96  metres,  and  the 
equatorial  radius,  6377397.6  metres;  find  the  two  radii,  and  their 
difference,  to  the  nearest  hundredth  of  a  mile,  1  metre  being  equal 
to  0.000621346  of  a  mile. 


DIWSION. 

SS3.  In  division  of  decimals  the  location  of  the  decimal 
point  in  the  quotient  depends  upon  the  following  principles : 

I.  If  one  decimal  number  in  the  fractional  form  be  divided  by 
another  also  in  the  fractional  form,  the  denominator  of  the  quotient 
must  contain  as  many  ciphers  as  the  number  of  ciphers  in  the  de- 
nominator of  the  dividend  exceeds  the  number  in  the  denominator 
of  the  divisor.     Therefore, 

II.  The  quotient  of  one  number  divided  by  another  in  the  deci- 
mal form  must  contain  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceed  the  number  in  the  divisor. 

1.  Divide  34.368  by  5.37. 


OPERATION. 


Analysis.     We  first  divide  as 
f\^7  ^  ^A  QAQ  r  a  1  ^^  whole  numbers  ;  then,  since  the 

09*99     ^     *  dividend  has  3  decimal  places  and 

the  divisor  2,  we  point  off  3  —  2 

'^  ^"*^  3=  1  decimal  place  in  the  quotient, 

_Z—-S.  (II).     The  correctness  of  the  work 

is  shown  in  the  proof,  where  the 

„„^^„  dividend  and  divisor  are  written  as 

PROOF. 

common  fractions.     For,  when  we 

S_4  3,S  8    y     1  0  0  __.    6  4  A  4  -  ,     ,     , 

i^oO    ^   o3T         10        ^-^        have  canceled  the  denominator  of 

the  divisor  from  the  denominator 
of  the  dividend,  the  denominator  of  the  quotient  must  contain  as 


I 


DIVISION.  ;[33 


many  ciphers  as  the  number  in  the  dividend  exceeds  those  in  the 
divisor. 

234:.    Hence  the  following 

Rule.     Divide  as  in  ichole  numherSy  and  from  the  right  hand 

of  the  quotient  point  off  as  mani/  places  for  decimals  as  the  decimal 

places  in  the  dividend  exceed  those  in  the  divisor. 

Notes. — 1.  If  the  number  of  figures  in  the  quotient  be  less  than  the  excess  of 
the  decimal  places  in  the  dividend  over  those  in  the  divisor,  the  deficiency  must 
be  supplied  by  prefixin«j  ciphers. 

2.  If  there  be  a  remainder  after  dividing  the  dividend,  annex  ciphers,  and 
continue  the  division  ;  the  ciphers  annexed  are  decimals  of  the  dividend. 

3.  The  dividend  should  always  contain  at  least  as  many  decimal  phices  as  tho 
divisor,  before  commencing  the  division  :  the  quotient  figures  will  then  be  inte- 
gers till  all  the  decimals  of  the  dividend  have  been  used  in  the  partial  dividends. 

4.  To  divide  a  decimal  by  10,  100, 1000,  etc.,  remove  the  point  as  many  places 
to  the  left  as  there  are  ciphers  on  the  right  of  the  divisor. 

EXAMPLES   FOR   PRACTICE. 

1.  Divide  9.61^8  by  3.46.  Ans.  2.78. 

2.  Divide  46.1975  by  54.35.  Ans.  .85. 

3.  Divide  .014274  by  .061.  Aiis.  .234. 

4.  Divide  .95£  by  4.76. 

5.  Divide  345.15  by  .075.  Ans.  4602. 

6.  Divide  .8  by  476.3.  Ans,  .001679+. 

7.  Divide  .0026  by  .003. 

8.  Divide  3.6  by  .00006.  Ans.  60000. 

9.  Divide  3  by  450. 

10.  Divide  75  by  10000. 

11.  Divide  4.36  by  100000. 

12.  Divide  .1  by  .12. 

13.  Divide  645.5  by  1000. 

14.  If  25  men  build  154.125  rods  offence  in  a  day,  how  much 
does  each  man  build  ? 

15.  How  many  coats  can  be  made  from  16.2  yards  of  cloth, 
allowing  2.7  yards  for  each  coat? 

16.  If  a  man  travel  36.34  miles  a  day,  how  long  will  it  take 
him  to  travel  674  miles  ?  Ans.  18.547+days. 

17.  How  many  revolutions  will  a  wheel  14.25  feet  in  cir«um- 
ference  make  in  going  a  distance  of  1  mile  or  5280  feet  ? 

12 


134 


DECIMALS. 


CONTRACTED    DIVISION. 

S^o.  To  obtain  a  given  number  of  decimal  places  in 
tlie  quotient. 

In  division,  the  products  of  the  divisor  by  the  several  quotient 
figures  maybe  contracted,  as  in  multiplication,  by  rejecting  at  each 
step  the  unnecessary  figures  of  the  divisor,  (220). 

1.  Divide  790.755197  by  32.4687,  extending  the  quotient  to 
two  decimal  places. 


FIRST   CONTRACTED 

METHOD. 

COMMON 

METHOD. 

S2.4687)  790.755197  ( 
649  4 

24.35 

32.4687  )  790.7  55198  (  24.35 
649  3  74 

1413 

141 3  811 

129  9 

129  8  748 

114 

115 

0639 

97 

97 

4061 

17 

17 

65787 

16 

16 

23435 

SECOND    CONTRACTED    METHOD. 

32.4687  )  790.755197 


53.42 


1413 

114 

17 

1 


1|42352 
Analysis.  In  the  first  method 
of  contraction,  we  first  compare  the 
3  tens  of  the  divisor  with  the  79 
tens  of  the  dividend,  and  ascertain 
that  there  will  be  2  integral  places 
in  the  quotient ;  and  as  2  decimal 
places  are  required,  the  quotient 
must  contain  4  places  in  all.  Then 
assu:7iing  the  four  left  hand  figures  of  the  divisor,  we  say  324G  is  con- 
tained in  7907,  2  times ;  multiplying  the  assumed  part  of  the  divisor 
by  2,  and  carrying  2  units  from  the  rejected  part,  as  in  Contracted 
Multiplication  of  Decimals,  we  have  6494  for  the  product,  which  sub- 
tracted from  the  dividend,  leaves  1413  for  a  new  dividend.  Now, 
since  the  next  quotient  figure  will  be  of  an  order  next  below  the 
former,  we  reject  one  more  place  in  the  divisor,  and  divide  by  324, 
obtaining  4  for  a  quotient,  1299  for  a  product,  and  114  for  a  new  divi- 
dend.    Continuing  this  process  till  all  the  figures  of  the  divisor  are 


CONTRACTED  DIVISION.  I35 

rejected,  we  have,  after  pointing  off  2  decimals  as  required,  24.35  for 
a  quotient.  Comparing  the  contracted  with  the  common  method,  we 
see  the  extent  of  the  abbreviation,  and  the  agreement  of  the  corres- 
ponding intermediate  results. 

In  the  second  method  of  contraction,  the  quotient  is  written  with 
its  first  figure  under  the  lowest  order  of  the  assumed  divisor,  and  the 
other  figures  at  the  left  in  the  reverse  order.  By  this  arrangement, 
the  several  products  are  conveniently  formed,  by  multiplying  each 
quotient  figure  by  the  figures  above  and  to  the  left  of  it  in  the  divisor, 
by  the  rule  for  contracted  multiplication,  (222),  and  the  remainders 
only  are  written  as  in  (112). 

3SG.    From  these  illustrations  we  derive  the  following 
Rule.     I.  Compare  the  highest  or  left  hand  figure  of  the  divisor 
with  the  units  of  like  order  in  the  dividend,  and  determine  hoiv 
Tnany  figures  will  he  required  in  the  quotient. 

II.  For  the  first  contracted  divisor,  take  as  many  significant 
figures  from  the  left  of  the  given  divisor  as  there  are  places  re- 
quired in  the  quotient;  and  at  each  subsequent  division  reject  one 
place  from,  the  right  of  the  last  preceding  divisor. 

III.  In  multijplying  hy  the  several  quotient  figures,  carry  from 
the  rejected  figures  of  the  divisor  as  in  contracted  multiplication. 

Notes.  —  1.  Supply  ciphers,  at  the  right  of  either  divisor  or  dividend,  when 
necessary,  before  commencing  the  work. 

2.  If  the  first  figure  of  the  quotient  is  written  under  tne  lowest  assumed  figure 
of  the  divisor,  and  the  other  figures  at  the  left  in  the  inverted  order,  the  several 
products  will  be  formed  with  the  greatest  convenience,  by  simply  multiplying 
each  quotient  figure  by  the  figures  above  and  to  the  left  of  it  in  the  divisor. 


EXAMPLES    FOR   PRACTICE. 

1.  Divide  27.3782  by  4.3267,  extending  the  quotient  to  3  deci- 
mal places.  Ans.  6.328  =1=. 

2.  Divide  487.24   by  1.003675,  extending  the  quotient  to  2 
decimal  places. 

3.  Divide  8.47326  by  75.43,  extending  the  quotient  to  5  deci- 
mal places. 

4.  Divide  .8487564  hy  .075637,  extending  the  quotient  to  3 
decimal  places.  Ans.  11.221  zh. 


I^Q  DECIMALS. 

5.  Divide  478.325  by  1.43|,  extending  the  quotient  to  3  deci- 
mal places.  Ans.SS2M2±:. 

6.  Divide  8972.436  by  756.3452,  extending  the  quotient  to  4 
decimal  places. 

7.  Divide  1  by  1.007638,  extending  tlie  quotient  to  6  decimal 
places.  Ans.  .992425  =h. 

8.  Find  the  quotient  of  .95372843  divided  by  4  t. 736546,  true 
to  8  decimal  places. 

9.  Reduce  |f  ^f  to  a  decimal  of  4  places.         Ans.  .7448  =fc. 

CIRCULATING   DECIMALS. 

327.  Common  fractions  can  not  always  be  exactly  expressed  in 
the  decimal  form;  for  in  some  instances  the  division  will  not  be 
exact  if  continued  indefinitely. 

328.  A  Finite  Decimal  is  a  decimal  which  extends  a  limited 
number  of  places  from  the  decimal  point. 

239.  An  Infinite  Decimal  is  a  decimal  which  extends  an 
unlimited  number  of  places  from  the  decimal  point. 

230.  A  Circulating  Decimal  is  an  infinite  decimal  in  which 
a  figure  or  set  of  figures  is  continually  repeated  in  the  same  order; 
as  .3333  +  ,  or  .437437437  +  . 

231.  A  Eepetend  is  the  figure  or  set  of  figures  continually 
repeated.  When  a  repetend  consists  of  a  single  figure,  it  is  in- 
dicated by  a  point  placed  over  it;  when  it  consists  of  more  than 
one  fi«;ure,  a  point  is  placed  over  the  first,  and  one  over  the  last 
figure.  Thus,  the  circulating  decimals  .55555+  and  .324324324  +  , 
are  written,   5  and  .324. 

232.  A  repetend  is  said  to  be  expanded  when  its  figures  are 
continued  in  their  proper  order  any  number  of  places  toward  the 
right;  thus,  .24,  expanded  is  .2424  +  ,  or  .242424242  +  . 

233.  Similar  Repetends  are  those  which  begin  at  the  same 
decimal  place  or  order;  as  .37  and  .5,  .24  and  .375, 1.56  and  24.3. 

234.  Conterminous  Repetends  are  those  which  end  at  the 
same  decimal  place  or  order;  as  .75  and  1.53,  .567,  and  3.245. 

Note. — Two  or  more  repetends  are  Similar  and  Cofiterniiiious  when  they  begin 
and  end  at  the  same  deeimal  places  or  orders. 


I 


CIKCtJLATING  DECIMALS.  I37 


3«tS.  A  Pure  Circulating  Decimal  is  one  which  contains  no 
figures  but  the  repetend;  as  .7,  or  .704. 

^30.  A  Mixed  Circulating  Decimal  is  one  which  contains 
other  figureS;  called  finite  places,  before  the  repetend;  as  .54,  or 
.013245,  in  which  .5  and  .01  are  the  finite  places. 

PROPERTIES    OF    FINITE   AND    CIRCULATING   DECIMALS. 

337.    The  operations  in  circulating  decimals  depend  upon  the 

following  properties. 

NoTK. — ].  The  common  fractions  referred  to  are  understood  to  \>q  proper  frac- 
tions, in  their  lowest  terms. 

I.  Every  fraction  whose  denominator  contains  no  other  prime 
factor  than  2  or  5  will  give  rise  to  a  finite  decimal ;  and  the  num- 
ber of  decimal  places  will  be  equal  to  the  greatest  number  of  equal 
factors,  2  or  5,  in  the  denominator. 

For,  in  the  reduction,  every  cipher  annexed  to  the  numerator  mul- 
tiplies it  by  10,  or  introduces  the  two  prime  factors,  2  and  5,  and  also 
gives  1  decimal  place  in  the  result.  Hence  the  division  will  be  exact 
when  the  number  of  ciphers  annexed,  or  the  number  of  decimal 
places  obtained,  shall  be  equal  to  the  greatest  number  of  equal  factors, 
2  or  5,  to  be  canceled  from  the  denominater. 

II.  Every  fraction  whose  denominator  contains  any  other  prime 
factor  than  2  or  5,  will  give  rise  to  an  infinite  decimal. 

For,  annexing  ciphers  to  the  numerator  introduces  no  other  prime 
factors  than  2  and  5  ;  hence  the  numerator  will  never  contain  aU  the 
prime  factors  of  the  denominator. 

III.  Every  infinite  decimal  derived  from  a  common  fraction  is 
also  a  circulating  decimal;  and  the  number  of  places  in  the 
repetend  must  be  less  than  the  number  of  units  in  the  denominator 
of  the  common  fraction. 

For,  in  every  division,  the  number  of  possible  remainders  is  limited 
to  the  number  of  units  in  the  divisor,  less  1 ;  thus,  in  dividing  by  7, 
the  only  possible  remainders  are  1,  2,  3,  4,  5,  and  6.  Hence,  in  the 
reduction  of  a  common  fraction  to  a  decimal,  some  of  the  remainders 
must  repeat  before  the  number  of  decimal  places  obtained  equals  the 
number  of  units  in  the  denominator ;  and  this  will  cause  the  inter- 
mediate quotient  figures  to  repeat. 
12* 


138  DECIMALS. 

KoTES. — 2.  It  will  be  found  that  the  number  of  places  in  the  repetend  is  always 
equal  to  the  denominator  less  1,  or  to  some  factor  of  this  number.  Thus,  the 
repetend  arising  from  ^  has  7  —  1  =  6  places  ;  the  repetend  arising  from  §j  has 
^-^  =  5  places. 

3.  A  perfect  rcpeteud  is  one  which  consists  of  as  many  places,  less  1,  as  there 
are  units  in  the  denominator  of  the  equivalent  fraction. 

4.  If  the  denominator  of  a  fraction  contains  neither  of  the  factors  2  and  5,  it 
will  irive  rise  to  a  pure  repetend.  But  if  a  circulating  decimal  is  derived  from  a 
fraction  whose  denominator  contains  either  of  the  factors  2  or  5,  it  will  contain 
as  many  finite  places  as  the  greatest  number  of  equal  factors  2  or  5  in  the  de- 
nominator. 

lY.  If  to  any  number  we  annex  as  many  ciphers  as  there  are 
places  in  the  number,  or  more,  and  divide  the  result  by  as  many 
9's  as  the  number  of  ciphers  annexed,  both  the  quotient  and  re- 
mainder will  be  the  same  as  the  given  number. 

For,  if  we  take  any  number  of  two  places,  as  74,  and  annex  two 
ciphers,  the  result  divided  by  100  will  be  equal  to  74 ;  thus, 
7400  -^  100  =  74. 
Now  subtracting  1  from  the  divisor,  100,  will  add  as  many  units 
to  the  quotient,  74,  as  the  new  divisor,  99,  is  contained  times  in  74, 
(115,  II)  ;   thus, 

7400  -^  99  =  74  +  J j,  or  74^* ; 

that  is,  if  two  ciphers  be  annexed  to  74,  and  the  result  be  divided  by 
99,  both  quotient  and  remainder  will  be  74.  In  like  manner,  annex- 
ing three  ciphers  to  74,  and  dividing  by  999,  we  have 

74000  ^  999  =  74/^V ; 
and  the  same  is  true  of  any  number  whatever. 

Y.  Every  pure  circulating  decimal  is  equal  to  a  common  frac- 
tion whose  numerator  is  the  repeating  figure  or  figures,  and  whose 
denominator  is  as  many  9's  as  there  are  places  in  the  repetend. 

For,  if  we  take  any  fraction  whose  denominator  is  expressed  by 
some  number  of  9's,  as  ||,  then  according  to  the  last  property,  annex- 
ing two  ciphers  to  the  numerator,  and  reducing  to  a  decimal,  we  have 
U  =  24jJ4«  I^  lil^G  manner,  carrying  the  decimal  two  places  farther, 
.24JJ  =  .2424^'^  j  hence,  ||  —  24.  By  the  same  principle,  we  have 
I-  =.2 ;  ^V  =-  -01 ;  i:'^  =  -02  ;  ^h  =  -001 ;  f  ^|  =  .324  ;  and  so  on.  And 
it  is  evident  that  all  possible  repetends  can  thus  be  derived  from  frac- 
tions whose  numerators  are  the  repeating  figures,  and  whose  denomi- 
nators are  as  many  9's  as  there  are  repeating  figures. 


CIRCULATING  DECIMALS.  239 

KoTE  5. — It  follows  from  the  last  property,  that  any  fraction  from  which  a  pure 
repetend  can  be  derived  is  reducible  to  a  form  in  which  the  denominator  is  some 
number  of  9's ;  thus  -f %  =*  IuoSjjI  5  3f  =  JM-  '-^his  is  true  of  every  fraction 
whose  denominator  terminates  with  1,  3,  7,  or  9. 

YI.  Any  repetend  may  be  reduced  to  another  equivalent  repe- 
tend, by  expanding  it,  and  moving  either  the  second  point,  or 
both  points,  to  the  right;  provided  that  in  the  result  they  be  so 
placed  as  to  include  the  same  number  of  places  as  are  contained 
in  the  given  repetend,  or  some  multiple  of  this  number. 

For,  in  every  such  reduction,  the  new  repetend  and  the  given  repe- 
tend, when  expanded  indefinitely,  will  give  results  which  are  identical. 
Thus,  .536  =- .53G536,  or  .53G53G53G,  or  .5365,  or  .53653,  or  .5365365, 
or  .53653653653  ;  because  each  of  these  new  repetends,  when  ex- 
panded, gives  .53653653653653653653+. 


Note  6. — If  in  any  reduction,  the  new  repetend  should  not  contain  the  sam 
number  of  places,  or  some  multiple  of  the  same  number,  as  the  given  repetend 
we  should  not  have,  in  the  expansions,  the  same  fgnrea  repeated  in  the  sam 
oi-dnr. 


REDUCTION. 


CASE   I. 


2S8.  To  reduce  a  pure  circulating  decimal  to  a 
common  fraction. 

1.  Eeduce  .675  to  a  common  fraction. 

OPERATION.  Analysis.     Since  the  repetend  has  3 

^•7^ fi75  _  2  5  places,  we  take  for  the  denominator  of 

the  required   fraction   the   number  ex- 
pressed by  three  9's,  (237,    '^).     Hence, 

KuLE.  Omit  the  j)oints  and  the  decimal  sign^  and  write  the 
Jigures  of  the  repetend  for  the  numerator  of  a  common  fraction ^ 
and  as  many  9's  as  there  are  places  in  the  repetend  for  the  de- 
nominator, 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  .45  to  a  common  fraction.  Ans,   -^j. 

2.  E.educe  .66  to  a  common  fraction. 

3.  Reduce  .279  to  a  common  fraction.  Arts.  y\'j. 


Arts. 

*A1 
TTT* 

Ans 

12 
•    T3* 

Ans. 

4tV 

Ans. 

•  If- 

Ans. 

Vt"- 

Ans.  15^^^. 

140  DECIMALS. 

4.  Reduce  .423  to  a  common  fraction. 

5.  Reduce  .923076  to  a  common  fraction. 

6.  Reduce  .95121  to  a  common  fraction. 

7.  Reduce  4.72  to  a  mixed  number. 

8.  Reduce  2.297  to  an  improper  fraction. 

9.  Reduce  2.97  to  an  improper  fraction. 
Note.  —  According  to  237,  VF,  2.97  =  2.972. 

10.  Reduce  15.0  to  a  mixed  number. 

CASE    II. 

S39.   To  reduce  a  mixed  circulating  decimal  to  a 
common  fraction. 

1.  Reduce  .0756  to  a  common  fraction. 

OPERATION.  Analysts.  Since  .756  is  equal 

.0756  =  ^^Vo  =  tV^  **^  ^i^'  -^^^^  ^^^^^  ^^  A  of  Uh 

2.  Reduce  .647  to  a  common  fraction. 

OPERATION.  Analyws.    Reducing  the  finite 

(*A*j 6  4     1      7  pai't  and  tke  repetend  separately 

(34Q g4         ^  to  fractions,  we  have  yVj  +  ^h- 

=  — - — f-- — .  To   reduce   these   fractions  to  a 

900  900  1  •     . 

common  clenommator,   we   must 

^^  040  —  d4  +  ^  multiply  the  terms  of  the  first  by 

900  9  ;  but  the  numerator,  G4,  may 

647 64  be  multiplied  by  9  by  annexing 

^^        900  ^  cipher  and  subtracting  04  from 

583      .  the  result,   ffivina;  — ,    for 

the  first  fraction  reduced.     The 

^^}  numerator  of  the  sum  of  the  two 

.647  given  decimal.  fractions  will  therefore   be   640 

64   finite  figures.  —  64  +  7  =  583,  and  supplying 

'Too  the  common  denominator,  we  have 

IJJ.      In  the   second   operation, 


583 
900 


Ans.  the  intermediate  steps  are  omitted. 


Hence  the  followinf): 


Rule.     I.  From  the  given  circulathuj  decimal  subtract  the  finite 
party  and  the  remainder  will  he  the  required  numerator. 


^B  CIRCULATING   DECIMALS.  1^1 

^B      II.    Write  as  many  9's  as  there  are  figures  in  the  repetend,  with 
^^  as  many  ciphers  annexed  as  there  are  finite  decimal  figures^  for 
the  required  denominator. 

EXAMPLES   FOR   PRACTICE. 

i  1.  Reduce  .57  to  a  common  fraction.  Ans,  ||. 

2,  Reduce  .048  to  a  common  fraction.  Ans.  ^y^. 

3.  Reduce  .6472  to  a  common  fraction. 

I             4.  Reduce  .6590  to  a  common  fraction.  Ans.  ||. 

5.  Reduce  .04648  to  a  common  fraction.  Ans.  -^^^. 

6.  Reduce  .1004  to  a  common  fraction. 

7.  Reduce  .9285714  to  a  common  fraction.  Ans.  ||. 

18.  Reduce  5.27  to  a  common  fraction.  Ans.  ||. 

9.  Reduce  7.0126  to  a  mixed  number.  Ans.  7^1^. 

10.  Reduce  1.58231707  to  an  improper  fraction.  Ans.  |^|. 

11.  Reduce  2.029268  to  an  improper  fraction. 

I  CASE  III. 

S40.  To  make  two  or  more  repetends  similar  and 
conterminous. 

1.  Make  .47,  .53675,  and  .37234  similar  and  conterminous. 

OPERATION.  Analysis.     The  first  of 

the  given  repetends  begins 

.47        =  .47474747474747  ^  ^^  the  place  of  tenths,  the 

.53675  =  .53675675675675  v  Ans.      second  at  the  place  of  thou- 

.37234  =  .37234723472347  )  sandths,  and  the  third  at 

the  place  of  hundredths; 
and  as  the  points  in  any  repetend  cannot  be  moved  to  the  left  over 
the  finite  places,  we  can  make  the  given  repetends  similar,  only  by 
moving  the  points  of  at  least  two  of  them  to  the  right. 

Again,  the  first  repetend  has  2  places,  the  second  3  places,  and  the 
third  4  places ;  and  the  number  of  places  in  the  new  repetends  must 
be  at  least  12,  which  is  the  least  common  multiple  of  2,  3,  and  4. 
AVe  therefore  expand  the  given  repetends,  and  place  the  first  point  in 
eadi  new  repetend  over  the  third  place  in  the  decimal,  and  the  second 
point  over  the  fourteenth,  and  thus  render  them  similar  and  conter- 
minous.    Hence  the  following 


142  DECIMALS. 

KuLE.  I.  Expand  the  repetendsj  and  place  the  first  point  in 
each  over  the  same  order  in  the  decimal, 

II.  Place  the  second  point  so  that  each  new  repetend  shall  con- 
tain as  mavy  places  as  there  are  units  in  the  least  common  mul- 
tiple of  tJie  number  of  places  in  the  severed  given  repetends. 

Note. — Since  none  of  the  points  can  be  carried  to  the  left,  some  of  them  must 
be  carried  to  the  right,  so  that  each  repetend  shall  have  at  least  as  many  finite 
places  as  the  greatest  number  in  any  of  the  given  repetends. 

EXAMPLES    FOR    PRACTICE. 

1.  Make  .43,  .57,  .4567,  and  .5037  similar  and  conterminous. 

2.  Make  .578,  .37,  .2485,  and  04  similar  and  conterminous. 

3.  Make  1.34,4.56,  and  .341  similar  and  conterminous. 

4.  Make  .5674,  .34,  .247,  and  -67  similar  and  conterminous. 

5.  Make  1.24,  .0578,  .4,  and  .4732147  similar  and  conter- 
minous. 

6.  Make  .7,  .4567,  .24,  and  .346789  similar  and  conterminous. 

7.  Make  .8,  .*36,  .4857,  .34567,  and  .2784678943  similar  and 
conterminous. 

ADDITION  AND    SUBTRACTION. 

241.  The  processes  of  adding  and  subtracting  circulating  deci- 
Dials  depend  upon  the  following  properties  of  repetends : 

I.  If  two  or  more  repetends  are  similar  and  conterminous,  their 
denominators  will  consist  of  the  same  number  of  9's,  with  the 
game  number  of  ciphers  annexed.     Hence, 

II.  Similar  and  conterminous  repetends  have  the  same  denomi- 
nators and  consequently  the  same  fractional  unit. 

1.  Add  .54,  3.24  and,  2.785. 

oPERATiox.  Analysis.      Since    fractions   can  be 

54    =      54444  added  only  when  they  have   the  same 

c)  h:     o  c}'A'^\h  fractional  unit,  we  first  make  the  repe- 

.  "^  *^ .  ^    •  tends  of  the  given  decimals  similar  and 

2.785  ==  2.78527  conterminous.    We  then  add  as  in  finite 

6.57214  decimals,  observing,  how^ever,  that  the 

1  W'hich  we  carry  from  the  left  hand 

column  of  the  repetends,  must  also  be  added  to  the  right  hand  column  ; 

for  this  w^ould  be  required  if  the  repetends  were  further  expanded 

before  adding. 


CIRCULATING  DECIMALS.  ^43 


2.  From  7.4  take  2.  7852. 

OPERATION.  Analysis.     Since  one  fraction  can  be  subtracted 

7  J-iii  ^J^om  another  only  when  they  have  the  same  frac- 

'^  .  ^  .  tional  unit,  we  first  make  the  repetends  of  the  given 

•^'  *  "'^-'  decimals  similar  and  conterminous.     We  then  sub- 

4.6581  tract  as  in  finite  decimals;  observing  that  if  both 

repetends  were  expanded,  the  next  figure  in  the 
subtrahend  would  be  8,  and  the  next  in  the  minuend  4 ;  and  the  sub- 
traction in  this  form  would  require  1  to  be  carried  to  the  2,  giving  1 
for  the  right  hand  figure  in  the  remainder. 

^4^.  From  these  principles  and  illustrations  we  derive  the 
following 

Rule.  I.  Whe7i  necessari/^  make  the  repetends  similar  and  con- 
terminous. 

II.  To  add  ] — Proceed  as  in  Jinite  decimals^  observing  to  increase 
the  sum  of  the  right  hand  column  hy  as  many  units  as  are  carried 
from  the  left  hand  column  of  the  repetends. 

III.  To  subtract ;  —  Proceed  as  in  finite  decimals,  ohserving  to 
diminish  the  right  hand  figure  of  the  remainder  hy  1,  when  the 
repetend  in  the  subtrahend  is  greater  than  the  repetend  of  the 
minuend. 

lY.  Place  the  points  in  the  residt  directly  under  the  points  above. 

Note. — When  the  sum  or  difference  is  required  in  the  form  of  a  common  frac- 
tion, proceed  according  to  the  rule,  and  reduce  the  result. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  sum  of  2.4,  .32,  .56t,  7.0o6,  and  4.37  ? 

Ans.  14.7695877. 

2.  What  is  the  sum  of  .478,  .321,  .78564,  .32,  .5,  and  .4326  ? 

Ans.  2.8961788070698. 

3.  From  .7854  subtract  .59.  Ans.   .1895258. 

4.  From  57.0587  subtract  27.31.  Ans.  29.745o. 

5.  What  is  the  sum  of  .5,  .32,  and  .12  ?  Ans.  1. 

6.  What  is  the  sum  of  .4387,  .863,  .21,  and  .3554  ? 

7.  What  is  the  sum  of  3.6537,  3.135,  2.564,  and  .53  ? 

8.  From  .432  subtract  .25.  Ans.  .18243. 

9.  From  7.24574  subtract  2.634,  Ans.  4.3i. 


144 


DECIMALS. 


10.  From  .99  subtract  .433.  Ans,  .55656. 

11.  What  is  the  sum  of  4.638,  8.318,  .016,  .54,  and  .45? 


Ans.  13|f. 


12.  From  .4  subtract  .23. 


Alls. 


7 


MULTIPLICATION   AND   DIVISION. 


343.    1.  Multiply  2.428571  by  .063. 


OPERATION. 

2.428571  =  \^ 


.063 


7 

-—  TTTJ 


1'7    V 


7      17    

TTU  —  UJi  — 


.154  Atis, 


Analysis.  We  first  re- 
duce the  multiplicand  and 
multiplier  to  their  equiva- 
lent fractions,  and  obtain 
V"^  and  T?ff ;  then  V  X 


7 

TTUr 


2.  Divide  .475  by  .3753. 


OPERATION. 


475 
-9-5-5 


.475 


•3750=375  0 


Analysis.  The  dividend  re- 
duced to  its  equivalent  common 
fraction  is  4J4,  and  the  divisor 


^?|X  HIS  =  1.26  Ans. 


reduced  to  its  equivalent  com- 
mon fraction  is  f ^|J ;   and  ^J| 

_i.  s  "  /> ') 10 1  oi 

~  -S^^G  —  TT  —  ^'^^' 

944:.    From  these  illustrations  we  have  the  following 
Rule.     Reduce  the  given  numbers  to  common  fractions ;  then 
multiply  or  divide^  and  reduce  the  result  to  a  decimal. 


EXAMPLES   FOR  PRACTICE. 


1.  Multiply  3.4  by  .72. 

2.  Multiply  .0432  by  18. 

3.  Divide  .154  by  .2. 

4.  Divide  4.5724  by  .7. 

5.  Multiply  4.37  by  .27: 

6.  Divide  56.6  by  137. 

7.  Divide  .428571  by  .54. 

8.  Multiply  .714285  by  .27. 

9.  Multiply  3.456  by  .425. 
10.  Divide  9.17045  by  3.36. 


Ans.  2A72, 

Ans.  .7783. 

Ans.  .693. 

An^.  5.8793. 

Ans.  1.182. 

Ans.  .41362530. 

Ans.  .7857142. 

Ans.  .194805. 

Ans.  1.4710037. 

Alls.  2.72637. 


11.  Multiply  .24  by  .57.      Ans.  .1395775941230486685032. 


UNITED  STATES  MONEY.  ^45 


UNITED  STATES  MONEY. 

S4«S.  By  Act  of  Congress  of  August  8,  1786,  the  dollar  was 
declared  to  be  the  unit  of  Federal  or  United  States  Money;  and 
the  subdivisions  and  multiples  of  this  unit  and  their  denomina- 
tions, as  then  established,  are  as  shown  in  the 

TABLE. 

10  mills  make  1  cent. 
10  cents      "      1  dime. 
10  dimes     "      1  dollar. 
10  dollars  "      1  eagle. 

34:0.    By  examining  this  table  we  find 

1st.  That  the  denominations  increase  and  decrease  in  a  tenfold 
ratio. 

2d.  That  the  dollar  being  the  unit,  dimes,  cents  and  mills  are 
respectively  tenths,  hundredths  and  thousandths  of  a  dollar. 

3d.  That  the  denominations  of  United  States  money  increase 
and  decrease  the  same  as  simple  numbers  and  decimals. 

Hence  we  conclude  that 

I.  United  States  money  may  he  expressed  according  to  the.  deci- 
mal system  of  notation. 

II.  United  States  money  may  he  added,  suhtracted,  multiplied 
and  divided  in  the  same  manner  as  decimals, 

NOTATION    AND    NUMERATION. 

247.  The  character  $  before  any  number  indicates  that  it 
expresses  United  States  money.     Thus  ?75  expresses  75  dollars. 

948.  Since  the  dollar  is  the  unit,  and  dimes,  cents  and  mills 
are  tenths,  hundredths  and  thousandths  of  a  dollar,  the  decimal 
point  or  separatrix  must  always  be  placed  before  dimes.  Hence, 
in  any  number  expressing  United  States  money,  the  first  figure  at 
the  right  of  the  decimal  point  is  dimes,  the  second  figure  is  cents, 
the  third  figure  is  mills,  and  it  there  are  others,  they  are  ten- 
thousandths,  hundred-thousandths,  etc.,  of  a  dollar.  Thus,  $8.3125 
13  K 


146  DECIMALS. 

expresses  8  dollars  3  dimes  1  cent  2  mills  and  5  tenths  of  a  mill 
or  5  ten-thousandths  of  a  dollar. 

S40.  The  denominations,  eagles  and  dimes,  are  not  regarded 
in  business  operations,  eagles  being  called  tens  of  dollars  and 
dimes  tens  of  cents.  Thus  $24.19  instead  of  being  read  2  eagles 
4  dollars  1  dime  9  cents,  is  read  24  dollars  19  cents.  Hence, 
practically,  the  table  of  United  States  money  is  as  follows : 

10  mills  make  1  cent. 
100  cents      "      1  dollar. 

@oO.  Since  the  cents  in  an  expression  of  United  States  money 
may  be  any  number  from  1  to  99,  the  first  two  places  at  the  right 
of  the  decimal  point  are  always  assigned  to  cents.  Hence,  when 
the  number  of  cents  to  be  expressed  is  less  than  10,  a  cipher 
must  be  written  in  the  place  of  tenths  or  dimes.  Thus,  7  cents  is 
expressed  $.07. 

Notes.  —  1.  The  half  cent  is  frequently  written  as  5  mills  and  vice  versd. 
Thus,  $.37i  =  $.375. 

2.  Business  men  frequently  write  cents  as  common  fractions  of  a  dollar. 
Thus,  $5.19  is  also  written  $5-J^9^,  read  5  and  ^1^9^  dollars. 

3.  In  business  transactions,  when  the  Jinal  result  of  a  computation  contains  5 
mills  or  more,  they  are  called  one  cent,  and  when  less  than  5  they  are  rejected. 
Thus,  $2,198  would  be  called  $2.20,  and  $1,623  would  be  called  $1.62. 

EXAMPLES    FOR   PRACTICE. 

1.  Write  twenty-eight  dollars  thirty-six  cents. 

A71S.  $28.36. 

2.  Write  four  dollars  seven  cents. 

3.  Write  ten  dollars  four  cents. 

4.  Write  sixteen  dollars  four  mills. 

5.  Write  thirty-one  and  one-half  cents. 

6.  Write  48  dollars  If  cents.  Ans.  $48.01 1. 

7.  Write  1000  dollars  1  cent  1  mill. 

8.  Write  3  eagles  2  dollars  5  dimes  8  cents  4  mills. 

9.  Write  6}  cents. 

10.  Eead  the  following  numbers  : 

$21.18  $10.01  $     .8125 

$164.05  $201,201  $15.08J 

$7.90  $5.37i  $96,005 


UNITED  STATES  MONEY.  147 

REDUCTION. 

QCil.    Since  $1  =  100  cents  =  1000  mills,  it  is  evident, 

1st,  That  dollars  may  be  changed  or  reduced  to  cents  by  an- 
nexing two  ciphers;  and  to  mills  by  annexing  three  ciphers. 

2d.  That  cents  may  be  reduced  to  dollars  by  pointing  oiF  two 
figures  from  the  right;  and  mills  to  dollars  by  pointing  off  three 
fii>ures  from  the  ri^ht. 

3d.   That  cents  may  be  reduced  to  mills  by  annexing  one  cipher. 

4th.  That  mills  may  be  reduced  to  cents  by  pointing  off  one 
figure  from  the  right. 

OPERATIONS    IN    UNITED    STATES    MONEY. 

2o9.  Since  United  States  Money  may  be  added,  subtracted, 
multiplied  and  divided  in  the  same  manner  as  decimals,  (346, 
II),  it  is  evident  that  no  separate  rules  for  these  operations  are 
required. 

EXAMPLES    FOR   PRACTICE. 

1.  Paid  $3475.50  for  building  a  house,  $310.20  for  painting, 
$1287.371  for  furniture,  and  $207.12 J  for  carpets;  how  much 
was  the  cost  of  the  house  and  furniture  ?  Ans,   $5280.20. 

2.  Bought  a  pair  of  boots  for  $4.62J,  an  umbrella  for  $1.75,  a 
pair  of  gloves  for  $.87  J,  a  cravat  for  $1,  and  some  collars  for 
$.62 J;  how  much  was  the  cost  of  all  my  purchases? 

3.  Gave  $150  for  a  horse,  $175.84  for  a  carriage,  and  $62J  for 
a  harness,  and  sold  the  whole  for  $390.37};  how  much  did  I 
gain?  Am.  $2,035. 

4.  A  man  bought  a  farm  for  $3800,  which  was  $190. 87J  less 
than  he  sold  it  for;  how  much  did  he  sell  it  for? 

5.  A  lady  bought  a  dress  for  $10f,  a  bonnet  for  $5},  a  veil  for 
$2f,  a  pair  of  gloves  for  $.87},  and  a  fan  for  $|.  She  gave  the 
shopkeeper  a  fifty  dollar  bill;  how  much  money  should  he  return 
to  her?  Ans.  $29,875. 

6.  A  farmer  sold  150  bushels  of  oats  at  $.37}  a  bushel,  and  4 
cords  of  wood  at  $3|  a  cord.    He  received  in  payment  84  pounds  of 


148  DECIMALS. 

sugar  at  6i  cents  a  pound,  25  pounds  of  tea  at  $|  a  pound,  2 
barrels  of  flour  at  $5.87 J,  and  the  remainder  in  cash;  how  much 
cash  did  he  receive?  Ans.  §39.125. 

7.  A  speculator  bought  264.5  acres  of  land  for  $6726.  He 
afterward  sold  126.25  acres  for  $31}  an  acre,  and  the  remainder 
for  $33.75  an  acre;  how  much  did  he  g^in  by  the  transaction? 

8.  A  merchant  going  to  New  York  to  purchase  goods,  had 
$11000.  He  bought  40  pieces  of  silk,  each  piece  containing  28  J 
yards,  at  $.80  a  yard;  300  pieces  of  calicoes,  with  an  average 
length  of  29  yards,  at  11}  cents  a  yard;  20  pieces  of  broadcloths, 
each  containing  36.25  yards,  at  $3,875  a  yard;  112  pieces  of 
sheeting,  each  containing  30.5  yards,  at  $.06}  a  yard.  How 
much  had  he  left  with  which  to  finish  purchasing  his  stock  ? 

A71S,  $6064.62|. 

9.  If  139  barrels  of  beef  cost  $2189.25,  how  much  will  1 
barrel  cost?  Ans.  $15.75. 

10.  If  396  pounds  of  hops  cost  $44,748,  how  much  are  they 
worth  per  pound  ?  Ans.  $.113. 

11.  Bought  lOf  cords  of  wood  at  $4 J  a  cord,  and  received  for 
it  7.74  barrels  of  flour;  how  much  was  the  flour  worth  per  barrel  ? 

12.  If  a  hogshead  of  wine  cost  $287.4,  how  many  hogsheads 
can  be  bought  for  $4885.80  ?  Ans.  17. 

13.  A  butcher  bought  an  equal  number  of  calves  and  sheep  for 
$265 ;  for  the  calves  he  paid  $3f  a  head,  and  for  the  sheep  $2|- 
a  head;  how  many  did  he  buy  of  each  kind  ?  Ans.  40. 

14.  If  128  tons  of  iron  cost  $9632,  how  many  tons  can  be 
bought  for  $1730.75  ?  Ans,  23. 

15.  If  125  bushels  of  potatoes  cost  $41.25,  how  many  barrels, 
each  containing  2  J  bushels,  can  be  bought  for  $112.20  ? 

16.  A  grocer  on  balancing  his  books  at  the  end  of  a  month, 
found  that  his  purchases  amounted  to  $2475.36,  and  his  sales  to 
i?1936.40 ;  and  that  the  money  he  now  had  was  but  |  of  what  he 
had  at  the  beginning  of  the  month;  how  much  money  had  he  at 
the  beginning  of  the  month  ?  Ans.  $1347.40. 

17.  A  person  has  an  income  of  $3200  a  year,  and  his  expenses 
are  $138  a  month ;  how  much  can  he  save  in  8  years  ? 


UNITED  STATES  MONEY.  ;149 

18.  Sold  120  pieces  of  cloth  at  §45  J  a  piece,  and  gained  thereby 
$1026  ;  how  much  did  it  cost  by  the  piece  ?  Ans.  $37.20. 

19^  A  flour  merchant  paid  $3088.25  for  some  flour.  He  sold 
425  barrels  at  $Gi  a  barrel,  and  the  remainder  stood  him  in  $4.50 
a  barrel;  how  many  barrels  did  he  purchase  ?  Ans.  521. 

20.  If  36  engineers  receive  $6315.12  for  one  month's  work, 
how  many  engineers  will  $21927.50  pay  for  one  month  at  the  same 
rate?  Ans.  125. 

21.  A  person  having  $1378.56,  wishes  to  purchase  a  house 
worth  $2538,  and  still  have  $750  left  with  which  to  purchase  fur- 
niture; how  much  more  money  must  he  have?   A71S.  $1909.44. 

22.  A  mechanic  earns  on  an  average  $1.87i  a  day,  and  works  22 
days  per  month.  If  his  necessary  expenses  are  $2 5  J  a  month, 
how  many  years  will  it  take  him  to  save  $1116,  there  being  12 
months  in  a  year?  Ans.  6  years 

23.  Bought  27.5  barrels  of  sugar  for  $453.75^  and  sold  it  at  a 
profit  of  $3.62 J  a  barrel;  at  what  price  per  barrel  was  it  sold  ? 

24.  A  man  expended  $70.15  in  the  purchase  of  rye  at  $.95  a 
bushel,  wheat  at  $1.37  a  bushel,  and  corn  at  $.73  a  bushel,  buying 
the  same  quantity  of  each  kind ;  how  many  bushels  in  all  did  he 
purchase  ?  Am    69  bushels. 

25.  A  farmer  bought  a  piece  of  land  containing  375  J  acres,  at 
$22i  per  acre,  and  sold  i  of  it  at  a  profit  of  $1032|;  at  what 
price  per  acre  was  the  land  sold  ?  Ans.  $27.75. 

26  If  3i  cords  of  wood  cost  $11  37i,  how  much  will  204  cords 
cost?  Ans.  $65.40f. 

27.  If  I  of  a  hundred  pounds  of  sugar  cost  $6|,  how  much 
can  be  bought  for  $46.75,  at  the  same  rate  ? 

Ans.  5.5  hundred  pounds. 

28.  A  man  sold  a  wagon  for  $62.50,  and  received  in  pa^^ment 
12i  yards  of  broadcloth  at  $3i  per  yard,  and  the  balance  in  cofl'ee 
at  12 J  cents  per  pound;  how  many  pounds  of  coffee  did  he  re- 
ceive? Ans.   175  pounds. 

29.  Bought  320  bushels  of  barley  at  the  rate  of  16  bushels  for 
$10.04,  and  sold  it  at  the  rate  of  20  bushels  for  $17 J;  how  much 
was  my  profit  on  the  transaction  ?  Ans.  $79.20. 

13* 


150  DECIMALS. 

PROBLEMS 
INVOLVING    THE    RELATION    OF   TRICE,    COST,    AND    QUANTITY. 

PROBLEM    I. 

3o3.   Given,  tlie  price  and  the  quantity,  to  find  the  cost. 

Analysis.  The  cost  of  3  units  must  be  3  times  the  price  of  1  unit ; 
of  8  units,  8  times  the  price  of  1  unit ;  of  f  of  a  unit,  |  times  the  price 
of  1  unit,  etc.     Hence, 

EuLE.     Multiply  the  price  of  one  hi/  the  quantift/. 

PROBLEM    II. 

254.   G  iven,  the  cost  and  the  quantity,  to  find  the  price. 

Analysis.  By  Problem  I,  the  cost  is  the  product  of  the  price  mul- 
tiplied by  the  quantity.  Now,  having  the  cost,  which  is  a  product, 
and  the  quantity,  Avhich  is  one  of  two  factors,  we  have  the  product 
and  one  of  two  factors  given,  to  find  the  other  factor.     Hence, 

lluLE.     Divide  the  cost  hy  the  quantify. 
PROBLEM   IIL 

2*15.  Given,  the  price  and  the  cost,  to  find  the  quantity. 

Analysis.     Pteasoning  as  in  Problem  II,  we  find  that  the  cost  is 
the  product  of  two  factors,  and  the  price  is  one  of  the  factors,     Hence, 
PtULE.      Divide  the  cost  hy  the  price. 

PROBLEM    IV. 

256.  Given,  the  quantity,  and  the  price  of  100  or 
1000,  to  find  the  cost. 

Analysis.  If  the  price  of  100  units  be  multiplied  by  the  number 
of  units  in  a  given  quantity,  the  product  will  be  100  times  the  required 
result,  because  the  multiplier  used  is  100  times  the  true  multiplier. 
For  a  similar  reason,  if  the  price  of  1000  units  be  multiplied  by  the 
number  of  units  in  a  given  quantity,  the  product  will  be  1000  times 
the  required  result.     These  errors  can  be  corrected  in  two  ways, 

1st.  By  dividing  the  product  by  100  or  1000,  as  the  case  may  be;  or, 

2d.  By  reducing  the  given  quantity  to  hundreds  and  decimals  of  a 
hundred,  or  to  thousands  and  decimals  of  a  thousand.     Hence, 


PROBLEMS  IN  UNITED  STATES  MONEY.  151 

Rule.     Multiply  the  price  hy  the  quantity  reduced  to  hundreds 

and  decimals  of  a  hundred ,  or  to  thousands  and  decimals  of  a 

thousand. 

Note.  —  In  business  transactions  the  Roman  numerals  C  and  M  are  com- 
monly used  to  indicate  hundreds  and  thousands,  where  the  price  is  by  the  100 
or  lO'OO. 

PROBLEM   Y. 

2o7.  To  find  tlie  cost  of  articles  sold  by  the  ton  of 
2000  pounds. 

Analysis.  If  the  price  of  1  ton  or  2000  pounds  be  divided  by  2, 
the  quotient  will  be  the  price  of  J  ton  or  1000  pounds.  We  then  have 
the  quantity  and  the  price  of  1000  to  find  the  cost.     Hence, 

E-ULE.  Divide  the  price  of  1  to7i  hy  2,  and  multiply  the  quo- 
tient hy  the  numher  of  pounds  expressed  as  thousandths. 

EXAMPLES    IN    THE    PRECEDING   PROBLEMS. 

1.  What  cost  187  barrels  of  salt,  at  §1.32  a  barrel? 

Ans.  $246.84. 

2.  What  cost  5  firkins  of  butter,  each  containing  70^  pounds, 
at  $-^-^g  a  pound  ?  Ans.  $66.09|. 

3.  If  the  board  of  a  family  be  §501.87-^  for  1  year,  how  much 
is  it  per  day  ?  Ans.  §1.37^. 

4.  At  $.10J-  a  dozen,  how  many  dozen  of  eggs  can  be  bought 
for  $18.48  ?  Ans.  176. 

5.  ^Vliat  is  the  value  of  140  sacks  of  guano,  each  sack  contain- 
ing 162 1  pounds,  at  S17|  a  ton  ?  Ans,  $201.906.|. 

6.  Wliat  will  be  the  cost  oi  3240  peach  trees  at  $161  per  hun- 
dred? Ans.  $534.60. 

7.  At  $66.44  a  ton,  what  \A\\  be  the  cost  of  842|  tons  of  rail- 
road iron?  Ans.  $55992.31. 

8.  A  gentleman  purchased  a  farm  of  325.5  acres  for  $10660^ ; 
how  much  did  it  cost  per  acre?  Ans.  $32.75. 

9.  What  will  be  the  cost  of  840  feet  of  plank  at  $1.94  per  C; 
and  1262  pickets  at  $12^  per  M  ?  Ans.  $32,071. 

10.  At  $11  a  bushel,  how  many  bushels  of  wheat  can  be  bought 
for  $37.68|  ?  Ans.  25^  bushels. 


152  DECIMALS. 

11.  What  will  be  tlie  cost  of  2172  pounds  of  plaster,  at  $3,875 
a  ton?  Ans.  $4.208f 

12.  What  cost  I  of  456  bushels  of  potatoes  at  $.37^  a  bushel? 

13.  If  32^  barrels  of  apples  cost  $81.25,  what  is  the  price  per 
barrel?  Ans.  $2.50. 

14.  What  must  be  paid  for  24240  feet  of  timber  worth  $9.37i 
per  M.?  Ans.  $227}. 

15.  At  $5|  an  acre,  how  many  acres  of  land  can  be  bought  for 
$4234.37i?  Ans.  752^. 

16.  How  much  must  be  paid  for  972  feet  of  boards  at  $20.25 
per  M,  1575  feet  of  scantling  at  $2.87}  per  C,  and  8756  feet  of 
lath  at  $7}  per  M  ?  Ans.  $130,634}. 

17.  What  is  the  value  of  1046  pounds  of  beef  at  $4|  per  hun- 
dred pounds?  Ans.  $48.37f. 

18.  What  is  the  value  of  5840  pounds  of  anthracite  coai  at 
$4.7  a  ton,  and  4376  pounds  of  shamokin  coal  at  $5.25  a  ton  ? 

19.  At  $2.50  a  yard,  how  much  cloth  can  be  purchased  for  $2  ? 

20.  What  is  the  value  of  3700  cedar  rails  at  $5|  per  C  ? 

21.  How  much  is  the  freight  on  3840  pounds  from  New  York 
to  Baltimore,  at  $.96  per  100  pounds  ?  Ans.  $36,864. 

22.  What  is  the  value  of  9  pieces  of  broadcloth,  each  piece 
containing  271  yards,  worth  $2  J  a  yard  ?  ^^js.  $715.87}. 

23.  At  $.42  a  pound,  how  many  pounds  of  wool  may  be  bought 
for  $80,745?  Ans.  192}. 

24.  What  will  be  the  cost  of  327  feet  of  boards  at  $15}  per 
M;  672  feet  of  siding  at  $1.62}  per  C,  and  1108  bricks  at  $4} 
per  M?  Ans.  $20.69|. 

25  At  $  J  per  yard,  how  many  yards  of  silk  may  be  bought  for 
$151?  Ans.  18. 

26.  How  much  must  be  paid  for  the  transportation  of  18962 
pounds  of  pork  from  Cincinnati  to  New  York,  at  $10  a  ton? 

27.  If  15}  yards  of  silk  cost  $27.9,  what  is  the  price  per  yard  ? 

28.  What  cost  27860  railroad  ties  at  $125.38  per  thousand  ? 

29.  If  .7  of  a  ton  of  hay  cost  $134,  what  is  the  price  of  1  ton  ? 
SO.  What  is  the  value  of  720  pounds  of  hay  at  $12.75  a  ton, 

and  912  pounds  of  mill  feed  at  $15}  a  ton  ?         Ans.  $11,658. 


ACCOUNTS  AND  BILLS.  I53 

LEDGER  ACCOUNTS. 
S58.  A  Ledger  is  the  principal  book  of  accounts  kept  by  mer- 
chants and  accountants.  Into  it  are  brought  in  sununary  ibrui 
the  accounts  from  the  journal  or  day-book.  The  items  often  form 
long  columns,  and  accountants  in  adding  sometimes  add  more  than 
one  column  at  a  single  operation,  (@§j. 


ao 

(2-) 

(3.) 

(4.) 

?  42.17 

$   506.76 

§2371. 67 

§14763.84 

36.24 

194  32 

4571.84 

33276.90 

18.42 

427.90 

1690.50 

47061.39 

10.71 

173.26 

2037.69 

18242.76 

194.30 

71.32 

5094.46 

37364.96 

S47.16 

39.46 

876.54 

8410.31 

40.00 

152.60 

679.81 

5724.27 

12.94 

271.78 

155.48 

'56317.66 

86.73 

320.00 

4930.71 

81742.73 

271.19 

709.08 

3104.13 

22431.27 

103.07 

48.50 

1987.67 

40163.55 

500.50 

63.41 

5142.84 

32189.60 

7.59 

56.00 

276.30 

7063.21 

11.44 

410.10 

522.71 

3451.09 

81.92 

372.22 

3114  60 

9200.00 

110.10 

137.89 

1776.82 

1807.36 

107.09 

276.44 

7152.91 

56768.72 

207.16 

18.19 

9328.42 

63024.27 

97.20 

27.96 

472.19 

36180  45 

21.77 

157.16 

321.42 

90807.08 

150.15 

94.57 

2423.79 

28763.81 

427.26 

177.66 

1600.81 

87196.75 

316.42 

327.40 

5976.27 

4230  61 

114.61 

1132.16 

4318.19 

3719.84 

81.13 

876.57 

682,45 

1367.92 

37.50 

179.84 

3174.96 

8756.47 

ACCOUNTS  AND  BILLS. 

S^O.  A  Debtor,  in  business  transactions,  is  a  purchaser,  or  a 
person  who  receives  money,  goods,  or  services  from  another;  and 

^00.  A  Creditor  is  a  seller,  or  a  person  who  parts  with 
money,  goods,  or  services  to  another. 


;j[54  DECIMALS. 

36fl,    An  Account  is  a  registry  of  debts  and  credits. 

Notes. — 1.  An  ciccount  should  always  contain  the  names  of  both  the  parties  to 
the  transaction,  the  price  or  value  of  each  item  or  article,  and  the  date  of  the 
transaction. 

2.  Accounts  may  have  only  one  side,  which  may  be  either  debt  or  credit;  or  it 
may  have  two  sides,  debt  and  credit. 

369.  The  Balance  of  an  Account  is  the  difference  between 
the  amount  of  the  debit  and  credit  sides.  If  the  account  have 
only  one  side^  the  balance  is  the  amount  of  that  side. 

303.  An  Account  Current  is  a  full  copy  of  an  account, 
giving  each  item  of  both  debit  and  credit  sides  to  date. 

3154.  A  Bill,  in  business  transactions,  is  an  account  of  money 
paid,  of  goods  sold  or  delivered,  or  of  services  rendered,  with  the 
price  or  value  annexed  to  each  item. 

36^0  The  Footing  of  a  Bill  is  the  total  amount  or  cost  of  all 
the  items. 

Note. — A  bill  of  goods  bought  or  sold,  or  of  services  received  or  rendered  at 
a  single  transaction,  and  containing  only  one  date,  is  often  called  a  Bill  of  Par- 
cels;  and  an  account  current  having  only  one  side  is  sometimes  called  sl  Bill 
of  Items. 

3©@,  In  accounts  and  bills  the  following  abbreviations  are  in 
general  use : 

Dr.  for  debit  or  debtor ; 
Cr.  for  credit  or  creditor ; 
\.  or  acc^t  for  account; 

@  for  at  or  by ;  when  this  abbreviation  is  used  it  is  always 
followed  by  the  price  of  a  unit.  Thus,  3  yd.  cloth  @  $1.25,  sig- 
nifies 3  yards  of  cloth  at  $1.25' per  yard;  J  lb.  tea  @  $.75,  signi- 
fies }  pound  of  tea  at  $.75  per'pound. 

2S7.  When  an  account  current  or  a  bill  is  settled  or  paid, 
the  fact  should  be  entered  on  the  same  and  signed  by  the  creditor, 
or  by  the  person  acting  for  him.  The  ^\^.  or  bill  is  then  said  to 
be  receipted.  Accounts  and  bills  may  be  settled,  balanced  and 
receipted  by  the  parties  to  the  same,  or  by  agents,  clerks  or  attor- 
neys authorized  to  transact  business  for  the  parties. 


ACCOUNTS  AND  BILLS.  155 

EXAMPLES    FOR   PRACTICE. 

Required;  the  footings  and  balances  of  the  following  bills  and 
accounts. 

Bill :  receipted  hy  clcrh  or  agent. 

New  York,  July  10,  1860. 
Mr.  John  C.  Smith, 

Bo't  of  Hill,  Groves  &  Co., 


10  yd. 

Cassimere, 

@ 

$2,85 

16    " 

BIk.  Silk, 

1.12J 

72    " 

Ticking, 

.It 

42    ^' 

Bid.  Shirting, 

.16} 

12    " 

Pressed  Flannel, 

.40 

24J  « 

Scotch  Plaid  Prints, 

.56 

$82.03 
Rec^d  Payment, 

Hill,  Groves  &  Co., 

By  J.  W.  Hopkins. 
(2.) 
Bill :  receipted  hy  the  selling  party. 

Chicago,  Sept.  20,  1861. 
Chase  &  Kennard, 

Bo't  of  McDouGAL,  Fenton  &  Co., 
125    pr.    Boys'  Thick  Boots,          @  $1.25 
275     ^^        ^^     Calf       ^'               ''      1.75 
180     ^'        "     Kip        "               ''      1.12J 
210     "        ''      Brogans,  '               ''        .87J 
80     "     Women's  Fox'd  Ggiiters,  ''        .84 
95     ''           ''         Opera    Boots,  "        .90 
175     "           "         Enameled  "     "      1.06 
8  cases  Men's  Calf  Boots,            "    30.50 
8     ''     Congress  Pump  Boots,     ^^    35.75 
1     ''     Drill,  958  yd.,                   "        .lOJ 
40  gross  Silk  Buttons,  ''        .37 i  

§1828.79 
Rec^d  Payment y 

McDouGAL,  Fenton  &  Co. 


156 


Smith  &  Perkins, 


DECIMALS. 

(3.) 

BUI :  settled  hy  note. 

New  York,  May  4,  1860. 


Bo't  of  Kent,  Lowber  &  Co., 

40  chests  Green  Tea,              @  $27.60 

25      "     Black        «              '' 

19.20 

16      "      Imperial    «               " 

48.10 

12  sacks   Java  Coffee,             " 

17.75 

20  bbl.      Coffee  Sugar,  (A)    " 

26.30 

15     "       Crushed    "               " 

31.85 

36  boxes  Lemons,                   " 

3.87i 

42     "       Oranges,                   " 

4.12J 

25     <^       Raisins,                    " 

2.90 

?2961.60 

ec^d  Payment^  hy  note  at  6  mo. 

Kent,  Lowber  &  Co. 

(4-) 

Bill :  paid  hy  draft,  and  receipted  hy  Cleric. 

New  Orleans,  April  28,  1861. 
James  Carlton  &  Co. 

BoH  of  WiLLARD  &  Hale. 

150  bbl.  Canada  Flour,  @  $6.25 


275    " 

Genesee    " 

"      7.16 

170    " 

Philada.    « 

"     5.87J 

326  bu. 

Wheat, 

"      1.62i 

214    " 

Corn, 

«       .82 

300    " 

Barley, 

"       .91 

500    " 

Kye, 

"      1.06 

$5413.48 


Rec^d  Payment,  hy  Draft  on  N.  T. 


R.  S.  Clarke, 
For  Willard  &  Hale. 


ACCOUNTS  AND  BILLS. 


157 


(5.) 

Account  Current ;  not  balanced  or  settled. 

Philadelphia,  Nov.  1,  1860. 
Mr.  James  Cornwall, 

To  Dodge  &  Son,  Dr. 

April  15,  To  24  tons  Swedes  Iron,  @  $64.30     $ 


a        a 

"    15  cwt.  Eng.  Blister  Steel,    " 

10.25 

June  21, 

"      7  doz.  Hoes,  (Trowel  Steel) '^ 

7.78 

Aug.  10, 

"    25    "    Buckeye  Plows,         " 

8.45 

Oct.      3, 

"    14  Cross-cut  Saws, 

a 

16.12} 

a        a 

"    27  cwt.  Bar  Lead, 

a 

5.90 

u        u 

"    1840  lbs.  Chain, 

i( 

.09} 

Cr. 

$ 

May  25, 

By  20  M.  Boards,     @ 

$17.60 

. 

July  14, 

"  50  M.  Shingles,    " 

3.12J 

a      u 

"  42  M.  Plank,        " 

9.87J 

Sept.   5, 

''  Draft  on  New  York, 

$1000 

"    12, 

"   75  C.  Timber,     @ 

3.10 

a      a 

''  36  C.  Scantling,  " 

.87i 

$ 

Bal.  Due  Dodge  &  Son,        §356.61 


(6.) 


Dr. 


Account  Current  J  another  form  ;  balanced  by  note. 
Wm.  Richmond  <fc  Co.  in  a|c.  current  -with  Wood  <fc  Powell. 


Or. 


1850 
July 
Aug. 

Oct. 


Dec. 


To  896  pounds  butter, 
"  872       "       cheese, 
"  4811^    "        lard, 
"  509^    «        tallow, 
"81  dozen  egj-rs, 
"15  barrels  salt, 
"  41  hams,  963%  pounds,  .Viy^ 


:i8t0 

^.23 

Nov. 

.091^ 

.113^ 

■1\ 

Dec. 

1 

.131^ 
.16^, 

1    " 

■>•/ 

!l861 

.40     : 

Jan. 

V 

.12K 

1 

565 

25 

iiy  CI  barrels  apples,  $2.25 
'•  70  bushels  turnips,  .22 
"  56     *'  dried  apples,  .87  V^ 
"  31  drums  figs,  .68;^ 


■  Note  at  3  mo.  to  Bal. 


Wood  &  Powell. 


Boston,  Jan.  1, 1861. 


158  DECIMALS. 

'promiscuous  examples. 

1.  AYliat  cost  121  cords  of  wood  @  $4.87i  ?   Ans,  $61.54+. 

2.  At  $.37}  per  bushel,  how  many  barrels  of  potatoes,  each 
containing  2}  bushels,  can  be  purchased  for  $33.75?     Ans.  36. 

3.  If  36  boxes  of  raisins,  each  containing  36  pounds,  can  be 
bought  for  $97.20,  what  is  the  price  per  pound  ?      Ans.  $.075. 

4.  If  .625  of  a  barrel  of  flour  be  worth  $5.35,  what  is  a  barrel 
worth  ?  Ans.  $8.56. 

5.  What  is  the  difference  between  |  of  a  hundredth,  and  |  of 
a  tenth  ?  Ans.  .025. 

6.  What  is  the  product  of  814^^^  X  26||  correct  to  2  decimal 
places  ? 

7.  A  drover  bought  5  head  of  cattle  @  $75,  and  12  head  @ 
$68 ;  at  what  price  per  head  must  he  sell  them  to  gain  $118  on 
the  whole?  Ans.  $77. 

8.  If  1  pound  of  tea  be  worth  $.62},  what  is  .8  of  a  pound 
worth  ?  Ans.  $.5. 

9.  A  person  having  $27.96,  was  desirous  of  purchasing  an 
equal  number  of  pounds  of  tea,  coffee,  and  sugar;  the  tea  @ 
$.87},  the  coffee  @  $.18f,  and  the  sugar  @  $.10}.  How  many 
pounds  of  each  could  he  buy?  Ans.  24. 

10.  If  a  man  travel  13543.47  miles  in  365}  days,  how  far  does 
he  travel  in  i  of  a  day  ?  A71S.  32.445  .miles. 

11.  Bought  100  barrels  of  flour  @  $5.12},  and  250  bushels  of 
wheat  @  $1.06}.  Having  sold  75  barrels  of  the  flour  @  $6}, 
and  all  the  wheat  @  $lf,  at  what  price  per  barrel  must  the  re- 
mainder of  the  flour  be  sold,  to  gain  $221.87}  on  the  whole  invest- 
ment? ^?is.  $6.75. 

12.  If  114.45  acres  of  land  produce  4580.289  bushels  of  pota- 
toes, how  many  acres  will  be  required  to  produce  120.06  bushels  ? 

Ans.  3. 

13.  Divide  .0172J|  by  .03j^g,  and  obtain  a  quotient  true  to  4 
decimal  places.  Ans.  .5625=1=. 

14.  Divide  13.5  by  2}  hundredths.  Ans.  600. 

15.  A  man  agreed  to  build  59.5  rods  of  wall;  having  built  8.5 


PROMISCUOUS  EXAMPLES.  159 

Js  in  5  days,  how  many  days  will  be  required  to  finisli  tlie  wall 
at  the  same  rate?  A)is.  30  days. 

16.  A  farmer  exchanged  28}  bushels  of  oats  worth  ^.37  J  per 
bushel,  and  453  pounds  of  mill  feed  worth  §.75  per  hundred,  for 
12520  pounds  of  plaster;  how  mucb  was  the  plaster  worth  per 
ton?  .     Ans.  $2.25. 

17.  A  farmer  sold  to  a  merchant  3  loads  of  hay  weighing  re- 
spectively 1826,  1478,  and  1921  pounds,  at  $8.80  per  ton,  and 
281  pounds  of  pork  at  $5.25  per  hundred.  He  received  in  exchange 
31  yards  of  sheeting  @  $.09,  6}  yards  of  cloth  @  $4.50,  and  the 
balance  in  money;  how  much  money  did  he  receive? 

18.  If  35  yards  of  cloth  cost  $122.50,  what  will  be  the  cost  of 
29  yards?  Ans.  $101.50. 

19.  A  speculator  bought  1200  bushels  of  corn  @  $.56}.  He 
sold  375. J  bushels  @  $.60.  At  what  price  must  he  sell  the  re- 
mainder, to  gain  $168,675  on  the  whole  ? 

20.  If  a  load  of  plaster  weighing  1680  pounds  cost  $2,856,  how 
much  will  a  ton  of  2000  pounds  cost?  Ans.  $3.40. 

21.  If  .125  of  an  acre  of  land  is  worth  $15|,  how  much  are 
25.42  acres  worth  ? 

22.  A  farmer  had  150  acres  of  land,  which  he  could  have  sold 
at  one  time  for  $100  an  acre,  and  thereby  have  gained  $3900;  but 

.  after  keeping  it  for  some  time  he  was  obliged  to  sell  it  at  a  loss 
of  $2250.  How  much  an  acre  did  the  land  cost  him,  and  how 
much  an  acre  did  he  sell  it  for  ? 

23.  A  lumber  dealer  bought  212500  feet  of  lumber  at  $14,375 
per  M,  and  retailed  it  out  at  $1.75  per  C;  how  much  was  his 
whole  gain  ? 

24.  If  10  acres  of  land  can  be  bought  for  $545,  how  many 
acres  can  be  bought  for  $17712.50  ?  Ans.  325. 

25.  How  much  is  the  half  of  the  fourth  part  of  7  times  224.56  ? 

Ans.  196.49. 

26.  Sold  10450  feet  of  timber  for  $169.8125,  and  gained 
thereby  $39.1 8| ;  how  much  did  it  cost  per  C  ?        Ans.  $1.25. 

27.  If  $6,975  be  paid  for  .93  of  a  hundred  pounds  of  pork, 
how  much  will  1  hundred  pounds  cost  ? 


160  DECIMALS. 

28.  Three  hundred  seventy-five  thousandths  of  a  lot  of  dry 
goods,  valued  at  $4000,  was  destroyed  by  fire ;  how  much  would 
a  firm  lose  who  owned  .12  of  the  entire  lot  r  Ans.  ^180. 

29.  Reduce  (tt-^-tt)  X  4  of  l  to  a  decimal.         Ans.  .15. 

\4^       2|/        ^        ^ 

30.  If  7.5  tons  of  hay  are  worth  375  bushels  of  potatoes,  and 
1  bushel  of  potatoes  is  worth  8.33|,  how  much  is  1  ton  of  hay 
worth?  Ans.  ?16.GG§. 

31.  A  person  invested  a  certain  sum  of  money  in  trade;  at  the 
end  of  5  years  he  had  gained  a  sum  equal  to  84  hundredths  of  it, 
and  in  5  years  more  he  had  doubled  this  entire  amount.  How 
many  times  the  sum  first  invested  had  he  at  the  end  of  the  10 
years?  Ans.  3.68  times. 

32.  A  miller  paid  $54  for  grain,  y^^  of  it  being  barley  at  $.62  J 
per  bushel,  and  |  of  it  wheat  at  $1.87}  per  bushel;  the  balance 
of  the  money,  he  expended  for  oats  at  $.37}  per  bushel.  How 
many  bushels  of  grain  did  he  purchase  ?  Ans.  40. 

33.  A  merchant  tailor  bought  27  pieces  of  broadcloth,  each 
piece  containing  19i  yards,  at  $4.31}  a  yard;  and  sold  it  so  as  to 
gain  $381.87},  after  deducting  $9.62}  for  freight.  How  much 
was  the  cloth  sold  for  per  yard  ?  Ans.  $5.06}. 

34.  Bought  1356  bushels  of  wheat  @  $1.18},  and  736  bushels 
of  oats  @  $.41 ;  I  had  870  bushels  of  the  wheat  floured,  and  dis- 
posed of  it  at  a  profit  of  $235.87},  and  I  sold  528  bushels  of  the 
oats  at  a  loss  of  $13.62}.  I  afterward  sold  the'remainder  of  the 
wheat  at  $1.12}  per  bushel,  and  the  remainder  of  the  oats  at  $.31 
per  bushel;  did  I  gain  or  lose,  and  how  much? 

Ans.  I  gained  $171.07}. 

35.  The  sum  of  two  fractions  is  i||,  and  their  difierence  is 
J^l ;  what  are  the  fractions  ? 

36.  A  manufacturer  carried  on  business  for  3  years.  The  first 
year  he  gained  a  sum  equal  to  |  of  his  original  capital;  the  second 
year  he  lost  J  of  what  he  had  at  the  end  of  the  first  year ;  the 
third  year  he  gained  |  of  what  he  had  at  the  end  of  the  second 
year,  and  he  then  had  $28585.70.  How  much  had  he  gained  in 
the  3  years?  Ans.  $10594.70. 


CONTINUED  FRACTIONS.  161 


CONTINUED  FRACTIONS. 

368.  If  we  take  any  fraction  in  its  lowest  terms,  as  ||,  and 
divide  botli  terms  by  the  numerator,  we  shall  obtain  a  complex 
fraction,  thus : 

13__1 

54  ""4  +  2^ 
13 
Reducing  ^^3,  the  fractional  part  of  the  denominator,  in  the  same 

manner,  we  have, 

13_1 

54  ~  4  +  1 


6  +  l_ 
2 
Expressions  in  this  form  are  called  continued  fractions.     Hence, 
S69.   A  Continned  Fraction  is  a  fraction  whose  numerator  is 
1,  and   whose  denominator  is   a  whole   number  plus  a  fraction 
whose  numerator  is  also  1,  ^nd  whose  denominator  is  a  similar 
fraction,  and  so  on. 

270.  The  Terms  of  a  continued  fraction  are  the  several  sim- 
ple fractions  which  form  the  parts  of  the  continued  fraction. 
Thus,  the  terms  of  the  continued  fraction  given  above  are,  J,  i, 
and  J. 

CASE  I. 

371.  To  reduce  any  fraction  to  a  continued  fraction. 

1.  Reduce  ^||  to  a  continued  fraction. 

OPERATION.  Analysis.    We  divide  the  denominator, 

109        1  339,  by  the  numerator,  109,  and  obtain  3 

339        3+1  for  the  denominator  of  the  first  term  of 

91]^         the  continued  fraction.    Then  in  the  same 
-r^        manner  we  divide  the  last  divisor,  109,  by 
the  remainder,  12,  and  obtain  9  for  the  de- 
nominator of  the  second  term  of  the  continued  fraction.    In  like  man- 
ner we  obtain  12  for  the  denominator  of  the  final  term.     Hence  the 
following 

HuLE.      I.  Divide   the  greater   term,   hy  the   less,  and  the  last 
divisor  hy  the  last  remainder ,  and  so  on,  till  there  is  no  remainder. 
14*  L 


162  CONTINUED  FRACTIONS. 

II.  Write  1  yb?'  the  numerator  of  each  temn  of  the  continued 
fraction,  and  the  quotients  in  their  order  for  the  respective  denom- 
inators. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  jT^f^  to  a  continued  fraction. 

4  +  1 

2  +  1 

3  +  1_^ 
9* 

2.  Reduce  i|-U  to  a  continued  fraction. 

8.  Reduce  |f|f  Jf  ^^  ^  continued  fraction. 
4.  Reduce  -f^j  to  a  continued  fraction. 

CASE    II. 

^72.  To  find  the  several  approximate  values  of  a 
continued  fraction. 

An  Approximate  Value  of  a  continued  fraction  is  the  simple 
fraction  obtained  by  reducing  one^  t\YO^  tlireC;  or  more  terms  of  the 
continued  fraction. 

S73.  1.  Reduce  ^^^^3  to  a  continued  fraction^  and  find  its 
approximate  values. 

OPERATION. 

08     1 

r-r  ~  ^   ,   , ,  the  continued  fraction. 
1G3      4  -f-  r 

3+jL 

2  +  1 

5 


—  _,  1st  approx.  value. 
4 


1  •»  3 

=:  tt;,  2d      " 


r+T       4x3  +  1 


13' 


1 _1 3  x2  +  l_3X2  +  l_7.3d      u 

4  +  1 4  +  2 (4X3  +  l)X^  +  4      13X2  +  4      30 

3+1  3X2+1 


2 
1  7X5+3 


=  — -,  4th  « 


4+1  SO  X  5  +13       163' 

3TT 

2  +  1 
5 


CONTINUED  FRACTIONS.  1(53 

Analysis.  We  take  |,  the  first  term  of  the  continued  fraction, 
for  the  first  approximate  value.  Reducing  the  complex  fraction 
formed  by  the  first  two  terms  of  the  continued  fraction,  we  have  ^^ 
for  the  second  approximate  value.  In  like  manner,  reducing  the  first 
three  terms,  we  have  ^^  for  the  third  approximate  value.  By  exam^ 
ining  this  last  process,  we  perceive  that  the  third  approximate  value, 
■^Q,  is  obtained  by  multiplying  the  terms  of  the  preceding  approxima- 
tion, j%^,  by  the  denominator  of  the  third  term  of  the  continued  frac- 
tion, 2,  and  adding  the  corresponding  terms  of  the  first  approximate 
value.  Taking  advantage  of  this  principle,  we  multiply  the  terms  of 
^ff  by  the  4th  denominator,  5,  in  the  continued  fraction,  and  adding 
the  corresponding  terms  of  -j-\,  obtain  j^^,  the  4th  approximate  value, 
which  is  the  same  as  the  original  fraction.     Hence  the  following 

lluLE.  I.  For  the  first  ajyproxiinate  valuCj  talze  the  first  term 
of  the  continued  fraction. 

II.  For  the  second  approximate  valuer  reduce  the  complex  frac- 
tion formed  hy  the  first  two  terms  of  the  continued  fraction. 

III.  For  each  succeeding  approximate  value ,  mulflpltj  both  nu- 
mrrator  and  denominator  of  the  last  preceding  aj)p>roxlmation  hy 
the  next  denominator  in  the  continued  fraction  ^  and  add  to  the  cor- 
responding products  respectively  the  numerator  and  derioininator  of 
the  preceding  approximation. 

Notes.  —  1.  When  the  given  fraction  is  improper,  invert  it,  and  reduce  this 
result  to  a  continued  fraction;  then  invert  the  approximate  values  obtained 
therefrom. 

2.  In  a  series  of  approximate  values,  the  1st,  3d,  5th,  etc.,  are  greater  than 
the  given  fraction;  and  the  2d,  -Ith,  OLh,  etc.,  are  less  thim  the  given  fraction. 

EXAMPLES    FOR    PRACTICE. 

1.  Find  the  approximate  values  of  j%\. 

jLns.  jj,  77,  375^,  J55. 

2.  Find  the  approximate  values  of  -^W. 

Alii      "i        4         5         3  9         8_3 

3.  What  arc  the  first  three  approximate  values  of  ^tqW?  ^ 

4.  What  are  the  first  five  approximate  values  of  f  f  J  ? 

A  14         9         4  <^       _4_9_ 

JinS.    -3,    j-3,   ^-g,   j'27j?    158- 

5.  Reduce  p  to  the  form  of  a  continued  fraction,  and  find  the 
value  of  each  approximating  fraction. 


164  COMPOUND  NUMBERS. 


COMPOUND  NUMBEKS. 

^74.  A  Compound  Number  is  a  concrete  number  expressed 
in  two  or  more  denominations,  (lO). 

gy^.  A  Denominate  Fraction  is  a  concrete  fraction  whose 
integral  unit  is  one  of  a  denomination  of  some  compound  number. 
Thus,  I  of  a  day  is  a  denominate  fraction,  the  integral  unit  being 
one  day;  so  are  |  of  a  bushel,  |  of  a  mile,  etc.,  denominate  irac- 
tions. 

ST6.  In  simple  numbers  and  decimals  the  scale  is  uniform, 
and  the  law  of  increase  and  decrease  is  by  10.  But  in  compound 
numbers  the  scale  of  increase  and  decrease  from  one  denomination 
to  another  is  varying,  as  will  be  seen  in  the  Tables. 

MEASURES. 

^^y.  Measure  is  that  by  which  extent,  dimension,  capacity 
or  amount  is  ascertained,  determined  according  to  eome  fixed 
standard. 

Note.  —  The  process  by  which  the  extent,  dimension,  cnpacity,  or  amount  is 
ascertained,  is  called  Measurincj  ;  and  consists  in  comparing  the  thing  to  be 
measured  with  some  conventionul  standard. 

Measures  are  of  seven  kinds : 

1.  Length.  4.  Weight,  or  Force  of  Gravity. 

2.  Surface  or  Area.  5.  Time. 

3.  Solidity  or  Capacity.  6.  Angles. 

7.   Money  or  Value. 
The  first  three  kinds  may  be  properly  divided  into  two  classes — 
Measures  of  Extension,  and  Measures  of  Capacity. 

MEASURES  OF  EXTENSION. 

278.  Extension  has  three  dimensions  —  length,  breadth,  and 
thickness. 

A  Line  has  only  one  dimension  —  length. 

A  Surface  or  Area  has  two  dimensions  —  length  and  breadth. 


I 


MEASURES  OF  EXTENSION.  ^55 


A  Solid  or  Body  has  three  dimensions  —  length,  breadth,  and 
thickness. 

I.  Linear  Measure. 

S79.  Linear  Measure,  also  called  Long  Measure,  is  ueed  in 
measuring  lines  or  distances. 

The  unit  of  linear  measure  is  the  yard,  and  the  table  is  made 
up  of  the  divisors,  (feet  and  inches,)  and  the  multiples,  (rods, 
furlongs,  and  miles,)  of  this  unit. 

TABLE. 

12  inches  (in.)  make  1  foot, ft. 

3  feet  *'       1  yard, yd. 

5 J  yards,  or  16J  feet,  **       1  rod, rd. 

40  rods  *'       1  furlong, fur. 

8  furlongs,  or  320  rods,      '*      1  statute  mile, ..mi. 

UNIT    EQUIVALENTS. 

ft.  in. 

yd.  1  =  12 

rd.  1      rrr  3  =  36 

fur.  1    =  5J  :=        16J  =        198 

mi  1    =      40    =r      220    --=      660    =      7920 

1     =r    8     ==     320    =     1760    =    5280    =    63360 
Scale  —  ascending,  12,  3,  5  J,  40,  8;  descending,  8,  40,  5  J^  3,  12.  i 

The  following  denominations  are  also  in  use :  — 

3  barleycorns  make  1  inch,   {Z\ei;^Tt^^^^^^^ 

4  inches  "      1  hand,  I  j^'"^  in  measuring  the  height  of 

'  (  horses  directly  over  the  fore  feet. 
9  "  "1  span. 

21.888    ''  "      1  sacred  cubit. 

3       feet  "      1  pace. 

6  *'  "1  fathom,  used  in  measuring  depths  at  sea. 

1.15  statute    miles  "      1  geographic  mile,  |  "«^''  '"  measuring  dis- 

*    °     ^  '  ( tances  at  sea. 

3       geographic  *'    "      1  league. 
60  "  *  ^^  1   1  deo-ree  I  ^^  latitude  on  a  meridian  or  of 

GO.  16  statute  <<    "  j  o         |  longitude  on  the  equator. 

3GQ       degrees  "      the  circumference  of  the  earth. 

NoTKs. — 1.  For  the  purpose  of  measuring  cloth  and  other  goods  sold  by  the 
yard,  the  yard  is  divided  into  halves,  fourths,  eighths,  and  sixteenths.  The  old 
tnble  of  cloth  measure  is  practically  obsolete. 

2.  A  span  is  the  distance  that  can  be  reached,  spanned,  or  measured  between 
the  end  of  the  middle  finger  and  the  end  of  the  thumb.  Among  sailors  8  spans 
are  equal  to  1  fathom. 

3.  The  geographic  mile  is  ^^  of  -g-Jjy  or  ^  tVt^tj  ^^  ^^^  distance  round  the  center 
of  the  earth.     It  is  a  small  fraction  more  than  1.15  .statute  miles. 


166 


COMPOUND   NUMBERS. 


4.  The  length  of  a  degree  of  latitude  varies,  being  68.72  miles  at  the  equator, 
68.9  to  69.05  miles  in  middle  latitudes,  and  69.30  to  69.34  miles  in  the  polar 
regions.  The  mean  or  average  length,  as  stated  in  the  table,  is  the  standard 
recently  adopted  by  the  U.  S,  Coast  Survey.  A  degree  of  longitude  is  greatest 
at  the  equator,  where  it  is  09.16  miles,  and  it  gradually  decreases  toward  the 
poles,  where  it  is  0. 

surveyors'  linear  measure. 

380.  A  Gunter's  Chain,  used  by  land  surveyors^  is  4  rods 
or  66  feet  long,  and  consists  of  100  links. 

The  unit  is  the  chain,  and  the  table  is  made  up  of  divisors  and 
multiples  of  this  unit. 

TABLE. 

7.92  inches  (in.)  make  1  link, 1. 

''1  rod, rd. 

**      1  chain,  . . .  ch. 
**      1  mile,. . . .  mi. 


25 
4 

80 

links 
rods,  or 
chains 

66  feet, 

UNIT    EQl 

mi. 
1 

ch. 
1 

80 

rd. 
1 

4 

320 

1. 

1 

25 

100 

8000 


in. 

7.92 
198 
792 
63360 


Scale  — ascending,  7.92,  25,  4,  80;  descending,  80,  4,  25,  7.92. 

NoTK. — The  denomination,  rods,  is  seldom  used  in  chain  measure,  distances 
being  taken  in  chains  and  links. 

II.  Square  Measure. 
S81.    A  Square  is  a  figure  having  four  equal  sides  and  four 
equal  corners  or  right  angles. 

*I8^.  Area  or  Superficies  is  the  space  or  surface  included 
within  any  given  lines :  as^  the  area  of  a  square,  of  a  field,  of  a 
board,  etc. 

1  square  yard  is  a  figure  having  four 
sides  of  1  yard  or  3  feet  each,  as  shown 
in  the  diagram.  Its  contents  are  3x3 
=  9  square  feet.     Hence, 

Tlie  contents  or  area  of  a  square,  or 

of  any  other  figure  having  a  uniform 

length  and  a  uniform  breadth ,  is  found 

1  yd.  =  3  ft.  ^y  m^ultiplying  the  length  hy  the  hreadth. 


1  yd.  =  3  ft. 

9 

" 

feet 

30} 

<( 

yards 

40 

<( 

rods 

4 

roods 

640 

acres 

MEASURES  OF  EXTENSION.  Ig7 

Thus,  a  square  foot  is  12  inches  long  and  12  inches  wide,  and  the 
contents  are  12  X  12  =  144  square  inches.  A  board  20  feet  long  and 
10  feet  wide,  is  a  rectangle,  containing  20  x  10  =  200  square  feet. 

The  measurements  for  computing  area  or  surface  are  always 
taken  in  the  denominations  of  linear  measure. 

28S.  Square  Measure  is  used  in  computing  areas  or  sur- 
faces J  as  of  land,  boards,  painting,  plastering,  paving,  etc. 

The  unit  is  the  area  of  a  square  whose  side  is  the  unit  of 
length.  Thus,  the  unit  of  square  feet  is  1  foot  square;  of  square 
yards,  1  yard  square,  etc. 

TABLE. 

144     square  inches  (sq.  in.)  make  1  square  foot,. . .  .sq.  ft. 

1       "       yard,.,  .sq.  yd. 
1       **       rod,. .  .  .sq.  rd. 

1  rood, R. 

1  acre, .  A. 

1  square  mile,. . .  sq.  mi. 

UNIT    EQUIVALENTS. 

sq.  yd. 

1   = 

30}= 
1210  = 
4840  == 

1  =  640  =  2560  =  102400  =  3097600  =  27878400  =  4014489600 
Scale— ascending,  144,  9,  30},  40,  4,  640;  descending,  640,  4,  40, 
30},  9,  144. 

Artificers  estimate  their  work  as  follows : 

By  the  square  foot:  glazing  and  stone-cutting. 

By  the  square  yard :  painting,  plastering,  paving,  ceiling,  and 
paper-hanging. 

By  the  square  of  100  square  feet :  flooring,  partitioning,  roofing, 
slating,  and  tiling. 

Bricklaying  is  estimated  by  the  thousand  bricks,  by  the  square 
yard,  and  by  the  square  of  100  square  feet. 

;N'otes. — 1.  Tn  estimiiting  the  painting  of  moldings,  cornices,  etc.,  the  mea- 
suring-line is  carried  into  all  the  moldings  and  cornices. 

2.  in  estimating  brick-laying  by  either  the  square  yard  or  the  square  of  100 
feet,  the  work  is  understood  to  be  12  inches  or  1^  bricks  thick. 

.3.  A  thousand  shingles  are  estimated  to  cover  1  square,  being  laid  5  inches  to 
the  weather. 


sq.  rd. 

R. 

1  = 

A. 

1  = 

40  = 

1  = 

4  = 

160  = 

sq.  ft. 

sq. in. 

1  = 

144 

9  = 

1296 

272}== 

39204 

10890  = 

1568160 

43560  = 

6272640 

168 


COMPOUND  NUMBERS. 


surveyors'  square  measure. 
S§ 4.    This  measure  is  used  by  surveyors  in  computing  the 
area  or  contents  of  land. 


TABLE. 

625  square  links  (sq.  1.)                    make  1  pole, 

....  P. 

16  poles                                                  '*      1  square  chain, 

.  sq.  ch. 

10  square  chains                                   "      1  acre, 

....  A. 

640  acres                                                  "      1  square  mile, 

sq.  mi. 

36  square  miles  (6  miles  square)       "      1  township,  . . . 

...Tp. 

UNIT    EQUIVALENTS. 

P. 

sq.l. 

gq.  ch.                           1      = 

625 

A.                       1     ==               16     = 

10000 

sq.mi.                     1      =                10      ==                160      = 

100000 

Tp.            1     =        640    =        6400    =      102400    = 

64000000 

1     =     36     =     23040     =     230400     =     3686400     =    2304000000 

Scale  —  ascending,  625,  16,  10,  640,  36 ;  descending,  36,  640,  10, 

16,  625. 

Notes. — 1.   A  square  mile  of  land  is  also  called  a  section, 

2.  Canal  and  railroad  engineers  commonly  use  an  engineers'  chain,  which  con- 
sists of  100  links,  each  1  foot  long. 

3.  The  contents  of  land  are  commonly  estimated  in  square  miles,  acres,  and 
hundredths;  the  denomination,  rood,  is  rapidly  going  into  disuse. 

III.  Cubic  Measure. 
S8^.   A  Cube  is  a  solid,  or  body,  having  six  equal  square 
sides  or  faces. 

S80,    Solidity  is  the  matter  or  space  contained  within  the 
bounding  surfaces  of  a  solid. 

The  measurements  for  computing  solidity  are  always  taken  in 
the  denominations  of  linear  measure. 

If  each  side  of  a  cube  be  1  yard,  or  3 
feet,  1  foot  in  thickness  of  this  cube  will 
^      X      .iittM     contain  3x3x1=9  cubic  feet ;  and  the 
whole  cube  will  contain  3  X  3  X  3  =  27 
cubic  feet. 
II  '  '  IHHI         A  solid,   or  body,  may  have  the  three 

<^   \ lilW     dimensions   all   alike   or  all  different.     A 

body  4  ft.  long,  3  ft.  wide,  and  2  ft.  thick 
it,  _  I  ya,  contains  4  X  3  X  2  =  24   cubic  or  solid 

feet.     Hence  we  see  that 


Wr. 

tei 

fiiB 

■ 

\ 

|u/. 

1 

P 

.728    cubic  inches  (cu.  in.) 

27    cubic  feet 

40    cubic  feet  of  round  timber, 

or" 

50        '^           *'     hewn 

IG    cubic  feet 

8    cord  feet,  or 
128    cubic  feet 

241-  cubic  feet 

MEASURES  OF  EXTENSION.  Jgg 

The  cuhic  or  solid  contents  of  a  hody  are  found  hy  multiplymg 
the  length,  bread th,  and  thickness  together, 

QS7.  Cubic  Measure,  also  called  Solid  Measure,  is  used  in 
computing  the  contents  of  solids,  or  bodies;  as  timber,  wood 
stone,  etc. 

The  unit  is  the  solidity  of  a  cube  whose  side  is  the  unit  of 
length.  Thus,  the  unit  of  cubic  feet  is  a  cube  which  measures  1 
.foot  on  each  side ;  the  unit  of  cubic  yards  is  1  cubic  yard,  etc. 

TABLE. 

make  1  cubic  foot cu.  ft. 

"       1  cubic  yard cu.  yd. 

"       1  ton  or  load T. 

"       1  cord  foot cd.  ft. 

"       1  cord  of  wood Cd. 

ii       -j    I  perch  of  stone  ]  p  i^ 
I  or  masonry,      J 

Scale  —  ascending,  1728,  27.  The  other  numbers  are  not  in  a 
regular  scale,  but  are  merely  so  many  times  1  foot.  The  unit  equiva- 
lents, being  fractional,  are  consequently  omitted. 

Notes.  —  1.  A  cubic  yard  of  earth  is  called  a  load. 

2.  Railroad  and  transportation  companies  estimate  light  freight  by  the  space 
it  occupies  in  cubic  feet,  and  heavy  freight  by  weight. 

3.  A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains  1  cord; 
and  a  cord  foot  is  I  foot  in  length  of  such  a  pile. 

4.  A  perch  of  stone  or  of  masonry  is  16^  feet  long,  IJ  feet  wide,  and  1  foot 
high. 

5.  Joiners,  bricklayers,  and  masons,  make  an  allowance  for  windows,  doors, 
etc.,  of  one  half  the  openings  or  vacant  spaces.  Bricklayers  and  masons,  in  es- 
timating their  work  by  cubic  measure,  make  no  allowance  for  the  corners  of  the 
walls  of  houses,  cellars,  etc.,  but  estimate  their  work  by  the  girtj  that  is,  the 
entire  length  of  the  wall  on  the  outside. 

6.  Engineers,  in  making  estimates  for  excavations  and  embankments,  take  the 
dimensions  with  a  line  or  measure  divided  into  feet  and  decimals  of  a  foot.  The 
computations  are  made  in  feet  and  decimals,  and  the  results  are  reduced  to  cubic 
yards.  In  civil  engineering,  the  cubic  yard  is  the  unit  to  which  estimates  for 
excavations  and  embankments  are  finally  reduced. 

7.  In  scaling  or  measuring  timber  for  shipping  or  freighting,  -j^  of  the  solid 
contents  of  round  timber  is  deducted  for  waste  in  hewing  or  sawing.  Thus,  a 
log  that  will  make  40  feet  of  hewn  or  sawed  timber,  actually  contains  50  cubic 
feet  by  measurement;  but  its  market  value  is  only  equal  to  40  cubic  feet  of 
hewn  or  sawed  timber.  Hence,  the  cubic  contents  of  40  feet  of  round  and  5^ 
feet  of  hewn  timber,  as  estimated  for  market,  are  identical. 

15 


170  COMPOUND  NUxMBERS. 


MEASURES  OF  CAPACITY. 

# 

288.   Capacity  signifies  extent  of  room  or  space.' 
S89.   Measures  of  capacity  are  all  cubic  measures,  solidity  and 
capacity  being  referred  to  different  units,  as  will  be  seen  by  com- 
paring the  tables. 

Measures  of  capacity  may  be  properly  subdivided  into  two 
classes,  Measures  of  Liquids  and  Measures  of  Dry  Substances. 

I.    Liquid  Measure. 

290.  Liquid  Measure,  also  called  Wine  Measure,  is  used  in 
measuring  liquids ;  as  liquors,  molasses,  water,  etc. 

The  unit  is  the  gallon,  and  the  table  is  made  up  of  its  divisors 
and  multiples.      * 

TABLE. 

4    gills  (gi.)  make  1  pint, pt. 

2    pints  *'  1  quart, qt. 

4    quarts  "  1  gallon, gal. 

31 J  gallons  "  1  barrel, bbl. 

2    barrels,  or  63  gal.     "  1  hogshead, ..  hhd. 

UNIT   EQUIVALENTS. 


pt.     gi. 

qt.        1=4 

gal. 

1=2=    8 

bM.      1   = 

=   4  =   8  =   32 

hhd.     1  =  31J  = 

=  126  =  252  =  1008 

1  =  2  =  63   = 

=  252  =  504  =  2016 

Scale — ascending,  4,  2,  4,  31  J,  2;  descending,  2,  31J,  4,  2,  4. 

The  following  denominations  are  also  in  use : 

42  gallons  make  1  tierce. 

2  hogsheads,  or  126  gallons,       ''       1  pipe  or  butt. 
2  pipes  or  4  hogsheads,  "       1  tun. 

Notes. — 1.  The  denominations,  barrel  and  hogshead,  are  used  in  estimating 
the  capacity  of  cisterns,  reservoirs,  vats,  etc.  In  Massachusetts  the  barrel  is 
estimated  at  32  gallons. 

2.  The  tierce,  hogshead,  pipe,  butt,  and  tun  are  the  names  of  casks,  and  do 
not  express  any  fixed  or  definite  measures.  They  are  usually  gauged,  and  hnve 
their  capacities  in  gallons  mnrked  on  them.  Several  of  these  denominations  are 
8till  in  use  in  England,  (327—330). 


WEIGHTS, 


171 


BEER    MEASURE. 

t201.   Beer  Measure  is  a  species  of  liquid  measure  used  in 
measuring  beer,  ale,  and  milk. 
The  unit  is  the  gallon. 

TABLE. 

2    pints  (pt.)  make  1  quart, qt. 

4    quarts  "       1  gallon, gal. 

313    gallons  '*       1  barrel, bbl. 

IJ  barrels,  or  54  gallons,  **       1  hogshead, ..  hhd. 

UNIT    EQUIVALENTS. 

qt.  pt. 

gal.  1=2 

bbl.  1     =        4    =        8 

hhd.  1       =     36     ==     144    =    288 

1     =     IJ     =     54    =    216    =    432 

Scale  —  ascending,  2,  4,  36,  1};  descending,  IJ,  36,  4,  2. 

This  measure  is  not  a  standard ;  it  is  rapidly  falling  into  disuse. 

II.     Dry  Measure. 
SO^.   Dry  Measure  is  used  in  measuring  articles  not  liquid ; 
as  grain,  fruit,  salt,  roots,  ashes,  etc. 

The  unit  is  the  bushel,  of  which  all  the  other  denominations  in 
the  table  are  divisors. 

table. 

2  pints  (pt.)       make  1  quart, qt. 

8  quarts  "       1  peck, pk. 

4  pecks  *'      1  bushel, ..  bu.  or  bush, 

UNIT    EQUIVALENTS. 

qt.  pt. 

pfe.  1=2 

bu.                  1  =          8        =        16 

1        =        4  =        32        =        64 

Scale  —  ascending,  2,  8,  4 ;  descending,  4,  8,  2. 

WEIGHTS. 
303.  Weight  is  the  measure  of  the  quantity  of  matter  a  body 
contains,  determined  by  the  force  of  gravity. 

Note.  —  The  process  by  which  the  quantity  of  matter  or  the  force  of  gravity 
is  obtained  is  called  Weighing;  and  consists  in  comparing  the  thing  to  be 
weighed  with  some  conventional  standard. 


172  COMPOUND  NUMBERS. 

Three  scales  of  weight  are  used  in  the  United  States ;  namely, 
Troy,  Avoirdupois,  and  Apothecaries'. 

I.    Troy  Weight. 

994.  Troy  Weight  is  used  in  weighing  gold,  silver,  and 
jewels;  in  philosophical  experiments,  and  generally  where  great 
accuracy  is  required. 

The  unit  is  the  pound,  and  of  this  all  the  other  denominations 
in  the  table  are  divisors. 

TABLE. 

24  grains  (gr.)     make  1  pennyweight, ..  pwt.  or  dwt. 

20  pennyweights       "     1  ounce, oz. 

12  ounces  "     1  pound, ...lb. 

UNIT    EQU1YALENT8. 

pwt.  gr. 

oz.  1    =       24 

lb.  1    =      20    =      480 

1    =    12    -=    240    =    5760 

Scale  —  ascending,  24,  20,  12;  descending,  12,  20,  24. 

Note. — Troy  weight  is  sometimes  called  Goldsmiths'  Weight, 

II.     Avoirdupois  Weight. 

39«>.  Avoirdupois  Weight  is  used  for  all  the  ordinary  pur- 
poses of  weighing. 

The  unit  is  the  pound,  and  the  table  is  made  up  of  its  divisors 
and  multiples. 

TABLE. 

16  drams  (dr.)  make  1  ounce, oz. 

16  ounces  *'     1  pound, lb. 

100  lb.                                 "     1  hundred  weight, ..  cwt. 
20  cwt.,  or  2000  lbs.,       "     1  ton, T. 


UNIT  EQUIVALENTS. 

OZ. 

dr. 

lb.            1 

r= 

16 

cwt.      1  =    16 



256 

T. 

1  =   100  =   1600 

= 

25600 

1  = 

=  20  =  2000  =  32000 

= 

512000 

Scale —  ascending,  16,  16,  100,  20;  descending,  20,  100,  16,  15. 


WEIGHTS. 


173 


Note.  —  The  long  or  gross  ton,  hundred  weight,  and  quarter  were  formerly  in 
common  use;  but  they  are  now  seldom  used  except  in  estimating  English  goods 
at  the  U.  S.  custom-houses,  in  freighting  and  wholesaling  coal  from  the  Peuu- 
gylvania  mines,  and  ia  the  wholesale  iron  and  plaster  trade. 

LONG  TON  TABLE. 

28  lb.                           make  1  quarter,                 marked  qr. 

4  qr.    =     112  lb.       "      1  hundred  weight,       ''  cwt. 

20  cwt.  =  2240  lb.       "      1  ton,    ^                         '*  T. 
Scale  —  ascending,  28,  4,  20;  descending,  20,  4,  28. 

!S90«  The  weight  of  the  bushel  of  certain  grains  and  roots 
has  been  fixed  by  statute  in  many  of  the  States ;  and  these  statute 
weights  must  govern  in  buying  and  selling,  unless  specific  agree- 
ments to  the  contrary  be  made. 

TABLE   OP   AVOIRDUPOIS    POUNDS    IN    A   BUSHEL, 
As  prescribed  by  statute  in  the  several  States  named. 


COMMODITIES. 


Barley. 

Beans 

Blue  Grass  Seed 

Buckwheat 

Castor  Beans 

Clover  Seed 

Dried  Apples 

Dried  Peaches... 

Flax  Seed 

Hair 

Hemp  Seed....... 

Indian  Corn 

Ind.  Corn  in  ear 
Ind.  Corn  Meal. 
Mineral  Coal*... 

Oats 

Onions 

Peas 

Potatoes 

Kye 

l^ve  M.3al 

Sali^ 

Timothy  Seed... 

Wheat 

Wlieat  Bran 


48  48 
60:60 
I4I14 
50  52 

46 '46 
6o'60 

25 1 24 

56  56 

44 1 44 
56J56 


60  60 
54  56 


333^  ; 
57 


60 

50 

50 
45 

60 '60 
20 '20 


J 

1 
1 

he 

rt 

c 

ii 

i 

.2 

'Z. 

a. 

a 

X 

25 

1 

J 

1 

1 

c 

c 

1 

32 

46 

48 

4S 

48 
60 
14 

48 

48 
62 

48 

46 

47 

46 

45 

46 

42 

42 

52 
46 

50 

48 

42 

48 

46 

42 

60 

60 

60 

64 

60 

60 

60 

60 

28 

28 

24 

28 

28 

28 

28 

33 
56 

55 

55 

56 

28 

28 

11 

44 

56 

50 

56 
50 

56 

50 

52 

56 

58 

5G 

56 

56 

50 

56 

56 

!iO 

32 

30 

30 
52 

32 

32 

35 

57 

30 

30 

32 

60 

32 

34 

32 

50 

32 

30 
50 

GO 

60 

60 

60 

60 

60 

60 

60 

60 

32 

50 

56 
50 

56 

56 

56 

r.o 
45 

56 

56 

56 
44 

56 

56 

56 

"» 

56 

56 

60 

60 

60 

CO 

60 

20 

60 

60 

60 

60 

60 

60 

60 

)  60 
28 


40 
•  60 


*  In  Kentucky,  80  Ibs.of  Intuminous  coal  or  70  lbs.  of  cannul  coal  make  1  bushel. 

*  In  IVnnsylvania.  80  lbs.  coar(«e,  70  lb.«.  frround,  or  02  Ib.s.  fine  ealL  make  I  bushel;  and 
in  Tliinoi?,  50  Ib.s.  common  or  56  lb.«!.  fine  .'^alt  make  1  buslu'l. 

*  In  Maine,  64  lbs.  of  rata  bajja  turnips  or  beets  make  1  bushel. 

15* 


174  COMPOUND  NUMBERS. 

Notes.  —  1.  The  weight  of  a  barrel  of  flour  is  7  quarters  of  old,  or  long  ton 
weiglit. 

2.  The  weight  of  a  bushel  of  Indian  corn  and  rye,  as  adopted  by  most  of  the 
States,  and  of  a  bushel  of  salt  is  2  quarters  ;  and  of  a  barrel  of  salt  10  quarters, 
or  i  of  a  long  ton. 

The  following  denominations  are  also  in  use : 
5G  pounds  make  1  firkin  of  butter. 


100 
100 
196 
200 
280 


1  quintal  of  dried  salt  fish. 

1  cask  of  raisins. 

1  barrel  of  flour. 

1       "      "  beef,  pork,  or  fish. 

1       ''      "  salt  at  the  N.  Y.  State  salt  works. 


III.  Apothecaries'  Weight. 

^^7.  Apothecaries'  Weight  is  used  by  apothecaries  and  phy- 
sicians in  compounding  medicines ;  but  medicines  are  bought  and 
sold  by  avoirdupois  weight. 

The  unit  is  the  pound,  of  wdiich  all  the  other  denominations  in 
the  table  are  divisors. 

TABLE. 

20  grains  (gr.)  make  1  scruple, so.  or  9. 

3  scruples  "     1  dram, dr.  or  3;. 

8  drams  "     1  ounce, oz.  or  J. 

12  ounces  "     1  pound, lb.  or  lb. 

UNIT    EQUIVALENTS. 

PC.  gr. 

dr.                   1  =          20 

oz.                1     =          3  =          GO 

u,.  1    =      8    =      24    =      480 

1    =    12    =    9G    =    288    =    57G0 
Scale  — ascending,  20,  3,  8,  12;'  descending,  12,  8,  3,  20. 

apothecaries'  fluid  measure. 

^@§,  The  measures  for  fluids,  as  adopted  by  apothecaries  and 
physicians  in  the  United  States,  to  be  used  in  compounding  medi- 
cines, and  putting  them  up  for  market,  are  given  in  the  following 

TABLE. 

GO  minims,  {'^)  make  1  fluidrachm, f^. 

8  fluidrachms,         "  1  fluidounce, f^. 

IG  fluidounces,  '*  1  pint, 0. 

8  pints,  **  1  gallon, Cong. 


TIME. 


175 


UNIT    EQUIVALENTS. 

f^  1      =  60 

0.  1    =  8    =       480 

Cong.       1    =      16    =      128    =      7680 
1    =    8    =    128    =    2048    =    61440 
Scale  —  ascending,  60,  8,  16,  8 ;  descending,  8,  16,  8,  60. 

MEASURE  OF  TIME. 
2®0.    Time  is  tlie  measure  of  duration.     The  unit  is  the  day, 
and  tlie  table  is  made  up  of  its  divisors  and  multiples. 

TABLE. 

60  seconds  (sec.)  make  1  minute, min. 

60  minutes,  "  1  lioiir, h. 

24  hours,  "  1  day, da. 

7  days,  "  1  week, wk. 

365  days,  *'  1  common  year, yr. 

366  days,  "  1  leap  3^ear, •  • .  yr. 

12  calendar  months,  ''  1  year, .    yr. 

100  years,  *'      1  century, C. 

UNIT   EQUIVALENTS. 

min.  («ec. 

h.                    1  -=  60 

da.                1    ==           60  =  3600 

^k.                1    =        24    =        1440  =  86400 

1    =          7    :_      1G8    _      10080  =  604800 

yr.         nio.                (365    =    8760    =    525600  =  31536000 

1    =    12         =    |366    =    8784    =    527040  =  31622400 

Scale  —  ascending,  60,  60,  24,  7 ;  descending,  7,  24,  60,  60. 

The  calendar  year  is  divided  as  follows : — 

No.  of  month.  Season.  Names  of  months.        Abbreviations.  No  of  days. 

^  w*   f  1  January,  Jan.  31 

2  winter,      |  pgi^^uary,  Feb.  28  or  29 

3  [March,  Mar.  31 

4  Spring,       <  April,  Apr.  30 

5  I  May,  31 

6  [June,  Jun.  30 

7  Summer,    <  July,  31 

8  I  August,  ^^S'  31 

9  [September,  Sept.  30 

10  Autumn,    ■<  October,  Oct.  31 

11  (November,  Nov.  30 

12  Winter,         December,  Dec.  31 
Notes.  —  1.  In  most  business  transactions  30  days  are  called  1  month. 

2.  The  civil  day  begins  and  ends  at  12  o'clock,  midnight.     The  astronomi- 


170  "  COMPOUND  NUMBERS. 

cal  day,  used  by  astronomers  in  dating  events,  begins  and  ends  at  12  o'clock, 
noon.     The  civil  year  is  composed  of  civil  days. 

BISSEXTILE    OR    LEAP   YEAR. 

300.  The  period  of  time  required  by  the  sun  to  pass  from 
one  vernal  equinox  to  another,  called  the  vernal  or  tropical  year, 
is  exactly  365  da.  5  h.  48  min.  49.7  sec.  This  is  the  true  year, 
and  it  exceeds  the  common  year  by  5  h.  48  min.  49.7  sec. 

If  365  days  be  reckoned  as  1  year,  the  time  lost  in  the  calendar 
will  be 

In  1  yr.,        5  h.  48  min.  49.7  sec. 
*'  4  *'         23  ''  15     ''     18.8   ** 

The  time  thus  lost  in  4  years  will  lack  only  44  min.  41.2  sec.  of 
1  entire  day.     Hence, 

If  every  fourth  year  be  reckoned  as  leap  year,  the  time  gained  in 
the  calendar  will  be, 

In      4  yr.,  44  min.  41.2  sec. 

-100-    (==25X4    yr.)     18  h.  37     "     10      " 

The  time  thus  gained  in  100  years  will  lack  only  5  h.  22  min.  50 
see.  of  1  day.     Hence 

If  every  fourth  year  be  reckoned  as  leap  year,  the  centennial  years 
excepted,  the  time  lost  in  the  calendar  will  be, 

In  100  yr.,        5  h.  22  min.  50  sec. 
-  400  "         21  -  31     -     20    - 

The  time  thus  lost  in  400  years  lacks  only  2  h.  28  min.  40  sec.  of  1 
day.     Hence 

If  every  fourth  year  be  reckoned  as  leap  year,  3  of  every  4  cen- 
tennial years  excepted,  the  time  gained  in  the  calendar  will  be, 

In    400  yr.,        2  h.  28  min.  40  sec. 
"  4000  -         24  h.  46  min.  40  sec. 

The  following  rule  for  leap  year  will  therefore  render  the  calendar 
correct  to  within  1  day,  for  a  period  of  4000  years. 

I.  Every  year  that  is  exactly  divisible  by  4  is  a  leap  year, 
the  centennial  years  excepted ;  the  other  years  are  common  years. 

IT.  Every  centennial  year  that  is  exactly  divisible  by  400  is  a 
leap  year  ]  the  other  centennial  years  are  common  years. 

Notes.  —  1.  Julius  Caesar,  the  Roman  Emperor,  decreed  that  the  year  should 
consist  of  365  days  6  hours;  that  the  6  hours  should  be  disregarded  for  3  suc- 
cessive years,  and  an  entire  daj'  he  added  to  every  fourth  year.  This  day  was 
inserted  in  the  calendar  between  the  24th  and  25th  days  of  February,  and  is 
called  the  iutercalary  day.  As  the  Romans  counted  the  days  backward  from  the 
first  day  of  the  following  month,  the  24th  of  February  was  called  by  them  scjcto 


CmCULAR  MEASURE.  177 

cnleudaa  Jfortit,  the  sixth  before  the  calends  of  March.  The  intercalary  day 
which  followed  this  was  called  biHsejcto  culeiidoH  Martii;  hence  the  name 
hissejitile. 

2.  In  1582  the  error  in  the  calendar  as  established  by  Julius  CcEsar  had  in- 
creased to  10  days;  that  is,  too  much  time  had  been  reckoned  as  a  year,  until 
the  civil  year  was  10  «lays  behind  the  solar  year.  To  correct  this  error,  Pope 
Gregory  decreed  that  10  entire  days  should  be  stricken  from  the  calendar,  and 
that  the  day  following  the  3d  day  of  October,  1582,  should  be  the  14th.  This 
brought  the  vernal  equinox  at  March  21  —  the  date  on  which  it  occurred  in  the 
year  H25,  at  the  time  of  the  Council  (»f  Nice. 

3.  The  year  as  established  by  Julius  Caesar  is  sometimes  called  the  Jnlinn 
year;  and  the  period  of  time  in  which  it  was  in  force,  namely  from  46  years 
B.  C.  to  1582,  is  called  the  //»/m»*  Period. 

4.  The  year  as  established  by  Pope  Gregory  is  called  the  Gregorian  year,  and 
the  calendar  now  used  is  the  Gregorian  Calendar. 

5.  Most  Catholic  countries  adopted  the  Gregorian  Calendar  soon  after  it  was 
established.  Great  Britain,  however,  continued  to  use  the  Julian  Calendar  until 
1752.  At  this  time  the  civil  year  was  11  days  behind  the  solar  year.  To  cor- 
rect this  error,  the  British  Government  decreed  that  11  days  should  be  stricken 
from  the  calendar,  and  that  the  day  follovring  the  2d  day  of  September,  1752, 
should  be  the  14th. 

6.  Time  before  the  adoption  of  the  Gregorian  Calendar  is  called  Old  Style 
(0.  S),  and  since,  New  Style,  (N.  S.)  In  Old  Style  the  year  commenced  March 
25,  and  in  New  Style  it  cominences  January  1. 

7.  Russia  still  reckons  time  by  Old  Style,  or  the  Julian  Calendar;  hence  their 
dates  are  now  12  days  behind  ours. 

8.  The  centuries  are  numbered  from  the  commencement  of  the  Christian  era; 
the  months  from  the  commencement  of  the  year;  the  days  from  the  commence- 
ment of  the  month,  and  the  hours  from  the  commencement  of  the  day,  (12 
o'clock,  midnight.)  Thus,  May  2:!,  1860,  9  o'clock  A.  M.,  is  the  9th  hour  of  the 
23d  day  of  the  5th  month  of  the  60th  year  of  the  19th  century. 

MEASURE   OF   ANGLES. 

301.  Circular  Measure,  or  Circular  Motion,  is  used  princi- 
pally in  surveying,  navigation,  astronomy,  and  geography,  for 
reckoning  latitude  and  longitude,  determining  locations  of  places 
and  vessels,  and  computing  difference  of  time. 

Every  circle,  great  or  small,  is  divisible  into  the  same  number 
of  equal  parts :  as  quarters,  called  quadrants;  twelfths,  called  signs; 
360ths,  called  degrees,  etc.  Consequently  the  parts  of  different 
circles,  although  having  the  same  names,  are  of  different  lengths. 

The  unit  is  the  degree,  which  is  ^  J^  part  of  the  space  about  a 
point  in  any  plane.  The  table  is  made  up  of  divisors  and  multiples 
of  this  unit. 

TABLE. 

60  seconds  (^^)       make  1  minute,....^. 
60  minutes  "      1  degree, . .  . .  °. 

30  degrees  *'      1  sign, S. 

12  signs,  or  360°,       "      1  circle, C. 

M 


178  COMPOUND  NUMBERS. 

UNIT    EQUIVALENTS. 


1   =             60 

S              1   =         60   =         3600 

c. 

1   =     30   =     1800   =     108000 

1  = 

12   =   360   =   21600   =   1296000 

Scale —  ascending,  60,  60,  30,  12;  descending,  12,  30,  60,  60. 

Notes.  —  1.  Minutes  of  the  earth's  circumference  are  called  geographic  or 
nautical  miles. 

2.  The  denomination,  sic/ua,  is  confined  exclusively  to  Astronomy. 

3.  A  degree  has  no  fixed  linear  extent.  When  applied  to  any  circle  it  is  always 
j^-jy  part  of  the  circumference.  But,  strictly  speaking,  it  is  not  any  part  of  a 
circle. 

4.  90°  make  a  quadrant  or  right-angle; 
60°       "      **  sextant      "   J  of  a  circle. 


MISCELLANEOUS  TABLES. 
30S.  COUNTING. 

12  units  or  things  make  1  dozen. 

12  dozen  ''  1  gross. 

12  gross  ''  1  great  gross. 

20  units  "  1  score. 

3®3.    PAPER. 

24  sheets make 1  quire. 

20  quires  '*  1  ream. 

2  reams  "  1  bundle. 

5  bundles  "  1  bale, 

304.     BOOKS. 

The  terms  folio,  quarto,  octavo,  duodecimo,  etc.,  indicate  the 
number  of  leaves  into  which  a  sheet  of  paper  is  folded. 


A  sheet  folded  in     2  leaves 

is  called 

a  folio. 

A  sheet  folded  in    4  leaves 

a  quarto. 

or  4to. 

A  sheet  folded  in     8  leaves 

an  octavo, 

or  8vo. 

A  sheet  folded  in  12  leaves 

a  12mo. 

A  sheet  folded  in  16  leaves 

'   (( 

a  16mo. 

A  sheet  folded  in  18  leaves 

an  18mo. 

A  sheet  folded  in  24  leaves 

a  24mo. 

A  sheet  folded  in  32  leaves 

a  32mo. 

30^.     COPYING. 

72  words  make  1  folio  or  sheet  of  common  law. 
90       '*  "■      1     '*       "■       ''       "■  chancery. 


GOVERNMENT  STANDARDS.  179 

GOVERNMENT  STANDARDS 
OF    MEASURES     AND    WEIGHTS. 

S0&.  In  early  times,  almost  every  province  and  chief  city  had 
its  own  measures  and  weights;  but  these  were  neither  definite  nor 
uniform.  This  variety  in  the  weights  and  measures  of  different 
countries  has  always  proved  a  serious  embarrassment  to  commerce ; 
hence  the  many  attempts  that  have  been  made  in  modern  times  to 
establish  uniformity. 

The  English,  American,  and  French  Governments,  in  establish- 
ing their  standards  of  measures  and  weights,  founded  them  upon 
unalterable  principles  or  laws  of  nature,  as  will  be  seen  by  ex- 
amining the  several  standards. 

UNITED  STATES  STANDARDS. 

3I^T.  In  the  year  1834  the  U.  S.  Government  adopted  a  uni- 
form standard  of  weights  and  measures,  for  the  use  of  the  custom 
houses,  and  the  other  branches  of  business  connected  with  the 
General  Government.  Most  of  the  States  which  have  adopted 
any  standards  have  taken  those  of  the  General  Government. 

308.  The  invariahh  standard  unit  from  which  the  standard 
units  of  measure  and  weight  are  derived  is  the  day. 

Astronomers  have  proved  that  the  diurnal  revolution  of  the 
earth  is  entirely  uniform^  always  performing  equal  parts  of  a  revo- 
lution on  its  axis  in  equal  periods  of  duration. 

Having  decided  upon  the  invariable  standard  unit,  a  measure 
of  this  unit  was  sought  that  should  in  some  manner  be  connected 
with  extension  as  well  as  with  this  unit.  A  clock  pendulum 
whose  rod  is  of  any  given  length,  is  found  always  to  vibrate  the 
same  number  of  times  in  the  same  period  of  duration.  Having 
now  the  day  and  the  pendulum,  the  different  standards  hereafter 
given  have  been  determined  and  adopted. 

STANDARD    OF    EXTENSION. 

300.  The  JJ.  S.  standard  unit  of  measiires  of  extension  jV{\ict\\.QV 
linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or  36  inches, 


180  COMPOUND  NUMBERS. 

and  is  the  same  as  the  Imperial  standard  yard  of  Great  Britain. 
It  is  determined  as  follows  :  The  rod  of  a  pendulum  vibrating 
seconds  of  mean  time,  in  the  latitude  of  London,  in  a  vacuum,  at 
the  level  of  the  sea,  is  divided  into  391393  equal  parts,  and  360000 
of  these  parts  are  36  inches,  or  1  standard  yard.  Hence,  such  a 
pendulum  rod  is  39.1393  inches  long,  and  the  standard  yard  is 
lliiBi  of  the  length  of  the  pendulum  rod. 

'  STANDARDS    OF  CAPACITY. 

310*  The  U,  S.  standard  unit  of  liquid  measure  is  the  old 
English  wine  gallon,  of  231  cubic  inches,  which  is  equal  to 
8.33888  pounds  avoirdupois  of  distilled  water  at  its  maximum 
density;  that  is,  at  the  temperature  of  39.83°  Fahrenheit,  the  ba- 
rometer at  30  inches. 

311.  The  TJ,  S.  standard  unit  of  dry  measure  is  the  British 
Winchester  bushel,  which  is  18  J  inches  in  diameter  and  8  inches 
deep,  and  contains  2150.42  cubic  inches,  equal  to  77.6274  pounds 
avoirdupois  of  distilled  water,  at  its  maximum  density.  A  gallon, 
dry  measure,  contains  268.8  cubic  inches. 

Notes. — 1.  Grain  «and  some  other  commodities  are  sold  by  strinhen  measure, 
and  in  such  cases  the  "  measure  is  to  be  stricken  with  a  round  stick  or  roller, 
straight,  and  of  the  same  diameter  from  end  to  end." 

2.  Coal,  ashes,  marl,  manure,  corn  in  the  ear,  fruit  and  roots  are  sold  by  heap 
measure.  The  bushel,  heap  measure,  is  the  Winchester  bushel  heaped  in  the 
form  of  a  cone,  which  cone  must  be  19^  inches  in  diameter  (=  to  the  outside 
diameter  of  the  standard  bushel  measure,)  and  at  least  6  inches  high.  A  bushel, 
heap  measure,  contains  2747.7167  cubic  inches,  or  597.2067  cubic  inches  more 
than  a  bushel  stricken  measure.  Since  1  peck  contains  ii5^'*  =  637.605  cubic 
inches,  the  bushel,  heap  measure,  contains  59.6917  cubic  inches  more  than  5 
pecks.  As  this  is  about  1  bu.  1  pk.  If  pt.,  it  is  sufficiently  accurate  in  practice, 
to  call  5  pecks  stricken  measure  a  heap  bushel. 

.3.  A  standard  bushel,  stricken  measure,  is  commonly  estimated  at  2150.4 
cubic  inches.  The  old  English  standard  bushel  from  which  the  United  States 
Etand:ird  bushel  was  derived,  was  kept  at  Winchester,  England;  hence  the  name. 

4.  The  wine  and  drj'  measures  of  the  same  denomination  are  of  difFerentcapac- 
ities.       The  exact  and  the  relative  size  of  each  may  be  readily  seen  by  the  fol' 


31^.     COMPARATIVE    TABLE    OF    MEASURES    OF    CAPACITY. 

Cubic  in.  in        Cubic  in.  in         Cubic  in.  in        Cubic  in.  ia 
one  frallon.  one  quart.  one  pint.  ouejrill. 

Wine  measure, 231  57|  28  J  7  ^\ 

Dry  measure  (Jpk.,)..   2G8J  67i  ^^  Sf 

Note.    —    The  beer  gallon  of  282  inches  is  retained  in  use  onl.v  by  custom. 


GOVERNMENT  STANDARDS.  181 

STANDARDS    OF   WEIGHT. 

313.  It  has  been  found  that  a  given  volume  or  quantity  of 
distilled  rain  water  at  a  given  temperature  always  weighs  the  same. 
Hence,  a  cubic  inch  of  distilled  rain  water  has  been  adopted  as 
the  standard  of  weight. 

314.  The  U.  S.  standard  unit  of  weiglit  is  the  Troy  pound  of 
Ihe  Mint,  which  is  the  same  as  the  Imperial  standard  pound  of 
Great  Britain,  and  is  determined  as  follows :  A  cubic  inch  of  dis- 
tilled water  in  a  vacuum,  weighed  by  brass  weights,  also  in  a 
vacuum,  at  a  temperature  of  62°  Fahrenheit's  thermometer,  is 
equal  to  252.458  grains,  of  which  the  standard  Troy  pound  con- 
tains 5760. 

31«l*  The  U.  S.  Avoirdupois  pound  is  determined  from  the 
standard  Troy  pound,  and  contains  7000  Troy  grains.  Hence, 
the  Troy  pound  is  f  ^g§  =  j4|  of  an  avoirdupois  pound.  But 
the  Troy  ounce  contains  ^^^^  =  480  grains,  and  the  avoirdupois 
ounce  '''-Jg^  =  437.5  grains;  and  an  ounce  Troy  is  480  —  437.5 
=  42.5  grains  greater  than  an  ounce  avoirdupois.  The  pound, 
ounce,  and  grain.  Apothecaries'  weight,  are  the  same  as  the  like 
denominations  in  Troy  weight,  ihoi  only  difference  in  the  two 
tables  being  in  the  divisions  of  the  ounce. 

3i6«    COMPARATIVE    TABLE    OF   WEIGHTS. 

Troy.  Avoirdupois.  Apothecaries'. 

1  pound     =     5760  grains,     =     7000  grains.    =     5760  grains, 
1  ounce     =      480       ''  =      437.5    "        =       480       " 

175  pounds,   =       144  pounds.  =       175  pounds, 

STANDARD    SETS    OF   WEIGHTS    AND    MEASURES. 

ly.  A  uniform  set  of  weights  and  measures  for  all  the  States 
was  approved  by  Congress,  June  14,  1836,  and  furnished  to  the 
States  in  1842.     The  set  furnished  consisted  of 

A  yard. 

A  set  of  Troy  weights. 

A  set  of  Avoirdupois  weights. 


182  COMPOUND   NUMBERS. 

A  wine  gallon^  and  its  subdivisions. 
A  half  bushel,  and  its  subdivisions. 

«li8«    State  Sealers  of  AVeights  and  Measures  furnish  standard 
sets  of  weights  and  measures  to  counties  and  towns. 
A  Count}/  Standard  consists  of 

1.  A  large  balance,  comprising  a  brass  beam  and  scale  dishes, 
with  stand  and  lever. 

2.  A  small  balance,  with  a  drawer  stand  for  small  weights. 

3.  A  set  of  large  brass  weights,  namely,  50,  20,  10,  and  5  lb. 

4.  A  set  of  small  brass  weights,  avoirdupois,  namely,  4,  2,  and 
I  lb.,  8,  4,  2,  1,  J,  and  i  oz. 

5.  A  brass  yard  measure,  graduated  to  feet  and  inches,  and  the 
first  foot  graduated  to  eighths  of  an  inch,  and  also  decimally;  with 
a  graduation  to  cloth  measure  on  the  opposite  side ;  in  a  case. 

6.  A  set  of  liquid  measures,  made  of  copper,  namely,  1  gal.,  J 
gal.,  1  C[t.,  1  pt.,  J  pt.,  1  gi.;  in  a  case. 

7.  A  set  of  dry  measures,  of  copper,  namely,  J  bu.,  1  pk.,  }  pk. 
(or  1  gal.),  2  qt.  (or  J  gal.),  1  qt.;  in  a  case. 

ENGLISH  MEASURES  AND  WEIGHTS. 
GOVERNMENT    STANDARDS. 

310.  The  English  act  establishing  standard  measures  and 
weights,  called  '"  The  Act  of  Uniformity,"  took  effect  Jan.  1, 1826, 
and  the  standards  then  adopted,  form  what  is  called  the  Imperial 
Sijdcm. 

3^0.  The  Ivvarialle  Standard  Unit  of  this  system  is  the 
same  as  that  of  the  United  States,  and  is  described  in  the  Act  of 
Uniformity  as  follows :  "  Take  a  pendulum  which  will  vibrate 
seconds  in  London,  on  a  level  of  the  sea,  in  a  vacuum;  divide  all 
that  part  thereof  which  lies  between  the  axis  of  suspension  and 
the  center  of  oscillation,  into  oOlSOo  equal  parts;  then  will  10000 
of  those  parts  be  an  imperial  inch,  12  whereof  make  a  foot,  and 
36  whereof  make  a  yard." 


I 


ENGLISH  MEASURES  AND  WEIGHTS.  183 

STANDARD    OF    EXTENSION. 

«|^I,  The  English  Standard  Unit  of  Measures  of  Extension, 
whether  linear,  superficial,  or  solid,  is  identical  vrith  that  of  the 
United  States,  (SI®®). 

STANDARDS    OF    CAPACITY. 

392.  Tlie  imjoerial  Standard  Gallon,  for  liquids  and  all  dry 
substances,  is  a  measure  that  will  contain  10  pounds  avoirdupois 
weight  of  distilled  water,  weighed  in  air,  at  G2°  Fahrenheit,  the 
barometer  at  30  inches.     It  contains  277.274  cubic  inches. 

393.  The  Imperial  Standard  Bmlicl  is  equal  to  8  gallons  or 
80  pounds  of  distilled  water,  weighed  in  the  manner  above  de- 
scribed.    It  contains  2218.192  cubic  inches. 

STANDARDS    OF    WEIGHT. 

324.  The  Imperial  Standard.  Pound  is  the  pound  Troy, 
w^hich  is  identical  with  that  of  the  United  States  Standard  Troy 
pound  of  the  Mint,  (314.) 

329S.  The  Imperial  At:oirdupois  Pound  contains  7000  Troy 
grains,  and  the  Troy  pound  5760.  It  also  is  identical  with  the 
United  States  avoirduj^ois  pound. 

TxVBLES. 

326.  The  denominations  in  the  standard  tables  of  measures 
of  extension,  capacity,  and  weights,  are  the  same  in  Great  Britain 
and  the  United  States.  But  some  denominations  in  several  of  the 
tables  are  in  use  in  various  parts  of  Great  Britain  that  are  not 
known  in  the  United  States. 

These  denominations  are  retained  in  use  by  common  consent, 
and  are  recognized  by  the  English  common  law.  They  are  as  fol- 
lows : 

327.     MEASURES    OF   EXTENSION. 

18  inches  make  1  cubit. 

45  inches  or  1        '<      1     11 

5  quarters  of  the  standard  yard  J 

NoTR. — The  cubit  wns  originally  the  length  of  a  mnn's  forearm  and  hand;  or 
the  distance  from  the  elbow  to  the  end  of  the  middle  finger. 


184  COMPOUND  NUMBERS. 

3S8.     MEASURES    OF    CAPACITY. 
LIQUID    MEASURES. 

9  old  ale  gallons  make  1  firkin. 

4  firkins  "  1  barrel  of  beer. 

7J  Imperial    "  "  1  firkin. 

52|  Imperial  gallons  or  |  ..  ^  hogshead. 
G3  wine                            j 

70  Imperial  gallons  or  I  **  1  puncheon  or 

84  wine              "            j  '*  J  of  a  tun. 

2  hogsheads,  that  is     | 

105  Imperial  gallons  or  v  "  1  pipe. 
126  wine              "            ) 

2  pipes  *'  1  tun. 

Pipes  of  wine  are  of  diiferent  capacities,  as  follows : 

110  wine  gallons  make  1  pipe  of  Madeira. 

(  Barcelona, 
120      ''         "  ^     "       1  Vidonia,  or 

[  Teneriffe. 
130      "         "  1     "  Sherry. 

138      "         ''  1     "  Port. 

I4Q      u         u  1     <'       I  Bucellas,  or 

I  Lisbon. 

3^9.     DRY    MEASURE. 

8  bushels  of  70  pounds  each  make  1  quarter  of  wheat. 
36       "        heaped  measure,        "       1  chaldron  of  coal. 

Note. — The  quarter  of  wheat  is  5C0  pounds,  or  i  of  a  ton  of  2240  pounds. 

330.     WEIGHTS. 

8  pounds  of  butchers'  meat        make  1  stone. 
14        "        ''    other  commodities       "      1      "      or  |^  of  a  cwt. 

2  stone,  or  28  pounds  *'      1  todd  of  wool. 

70  pounds  of  salt  "      1  bushel. 

Note. — The  English  quarter  is  28  pounds,  the  hundred  weight  is  112  pounds, 
and  the  ton  is  20  hundred  weight,  or  2210  pounds. 


FRENCH  MEASURES  AND  WEIGHTS. 
GOVERNMENT    STANDARDS. 

331.  The  tables  of  standard  measures  and  weights  adopted 
by  the  French  Government  are  all  formed  upon  a  decimal  scale, 
and  constitute  what  is  called  the  French  Metrical  S^»fem, 


FRENCH  MEASURES  AND  WEIGHTS.         185 

33^.  Invariable  Standard  Unit.  The  French  metrical  sys- 
tem has,  for  its  unit  of  all  measures,  whether  of  length,  area, 
solidity,  capacity,  or  weight,  a  uniform  invariable  standard,  adopted 
from  nature  and  called  the  mitre.  It  was  determined  and  estab- 
lished as  follows :  a  very  accurate  survey  of  that  portion  of  the 
terrestrial  meridian,  or  north  and  south  circle,  between  Dunkirk 
and  Barcelona,  France,  was  made,  under  the  direction  of  Govern- 
ment, and  from  this  measurement  the  exact  length  of  a  quadrant 
of  the  entire  meridian,  or  the  distance  from  the  equator  to  the 
north  pole,  was  computed.  The  ten  millionth  part  of  this  arc  was 
denominated  a  mitre,  and  from  this  all  the  standard  units  of 
measure  and  weight  are  derived  and  determined. 

STANDARDS    OF    EXTENSION. 

333.    The  French  Standard  Linear  Unit  is  the  m^tre. 

33z|^  The  French  Standard  Unit  of  Area  is  the  Are,  which 
is  a  unit  10  metres  square,  and  contains  100  square  metres. 

33o*  The  French  Standard  Unit  of  Solidity  and  Capaciti/ 
is  ohe  Litre,  which  is  the  cube  of  the  tenth  part  of  the  metre. 

STANDARD    OF   WEIGHT. 

33G  The  French  Standard  Unit  of  Weight  is  the  Gramme, 
which  is  determined  as  follows :  the  weight  in  a  vacuum  of  a 
cubic  decimetre  or  litre  of  distilled  water,  at  its  maximum  density, 
was  called  a  IcUogramme,  and  the  thousandth  part  of  this  was 
called  a  Gramme,  and  was  declared  to  be  the  unit  of  weight. 

NOMENCLATURE    OF    THE    TABLES. 

337*  It  has  already  been  remarked,  (331 ),  that  the  tables  are 
all  formed  upon  a  decimal  sea  e.  The  names  of  the  multiples  and 
divisors  of  the  Government  standard  units  in  the  tables  are  formed, 
by  combining  the  names  of  the  standard  units  with  prefixes ;  the 
names  of  the  niultiples  being  formed  by  employing  the  prefixes 
deca,  (ten),  hecto,  (hundred),  Icilo,  (thousand),  and  myria,  (ten 
thousand),  taken  from  the  Greek  numerals ;  and  the  names  of  the 
divisors  by  employing  the  prefixes  deci,  (tenth),  centi,  (hundredth), 
16* 


186  COMPOUND  NUMBERS. 

mili,  (thousandth),  from  the  Latin  numerals.  Hence  the  name 
of  any  denomination  indicates  whether  a  unit  of  that  denomination 
is  greater  or  less  than  the  standard  unit  of  the  table. 

338.  I.  French  Linear  Measure. 

TABLE. 

10  millimetres  make  1  centimetre. 

10  centimetres  "  1  decimetre. 

10  decimetres  "  1  metre. 

10  metres  "  1  decametre. 

10  decametres  "  1  hectometre. 

10  hectometres  "  1  kilometre. 

10  kilometres  "  1  myriametre. 

Notes. — 1.  The  metre  is  equal  to  39.3685  inches,  the  standard  rod  of  brass 
on  which  the  former  is  measured  being  at  the  temperature  of  32°  Fahrenheit, 
and  the  English  standard  brass  yard  or  **  Scale  of  Troughton"  at  62°.  Hence,  a 
metre  is  equal  to  3.2807  feet  English  measure. 

2.  The  length  of  a  metre  being  39.3685  inches,  and  of  a  clock  pendulum 
vibrating  seconds  at  the  level  of  the  sea  in  the  latitude  of  London  39.1393 
inches,  the  two  standards  differ  only  .2292,  or  less  than  i  of  an  inch. 

339.  II.  French  Square  Measure. 

TABLE. 

100  square  metres,  or  centiares  (10  metres  square)  make  1  are. 

100  ares  (10  ares  square)  *'      1  hectoare. 

Note.  —  A  square  metre  or  centiare  is  equal  to  1.19589444  square  yards,  and 
an  are  to  119.589444  square  yards. 

349.  III.  French  Liquid  and  Dry  Measure. 

TABLE. 

10  decilitres  make  1  litre. 

10  litres  "  1  decalitre. 

10  decalitres  "  1  hectolitre. 

10  hectolitres  "  1  kilolitre. 

Notes. — 1.  A  litre  is  equal  to  61.53294  cubic  inches,  or  1.06552  quarts  of  a  U. 
S.  liquid  gallon. 

2.  A  table  of  Solid  or  Cubic  Measure  is  also  in  use  in  some  parts  of  France, 
although  it  is  not  established  or  regulated  by  government  enactments  or  decrees. 
The  unit  of  this  table  is  a  cubic  metre,  which  is  equal  to  61532.94238  cubio 
inches,  or  35.60934  cubic  feet.     This  unit  is  called  a  Stere. 

TABLE. 

10  decisteres  make  1  stere. 

10  steres  **       1  decastere. 


MONEY  AND  CURRENCIES.  137 

•I41,   lY.  French  Weight. 

TABLE. 

10  milligrammes  make  1  centigramme. 

10  centigrammes  "  1  decigramme. 

10  decigrammes  "  1  gramme. 

10  grammes  "  1  decagramme. 

10  decagrammes  "  1  hectogramme. 

10  hectogrammes  "  1  kilogramme. 

100  kilogrammes  "  1  quintal. 

ir\       •   i.  1  n        f  1  millier,  or 

10  quintals  1  i   x         ^  a 

^  [1  ton  01  sea  water. 

Notes. — 1.  A  gramme  is  equal  to  15.483159  Troy  grains. 
2.   A  kilogramme  is  equal  to  2  lb.  8  oz.  3  pwt.  1.159  gr.  Troy,  or  2  lb.  3  oz. 
4.1549  dr.  Avoirdupois. 

342.   Comparative  Table  of  the  United  States,  English, 
AND  French  Standard  Units  of  Measures  and  Weights. 

United  States.  English.  French. 

Extonsion,    Yd.  of  3  ft.,  or  3(5  in.  Same  as  U.  S.  Metre.  39.3685  in. 

r.,n.,n\tv     I  ^^'"t^  fJ'd.,  231  cu.  in.  Imp'l  gal.,  277.274  cu.  in.    Litre,  61.53294  cu.  in. 

^.ip.icuy,    I  winch'r  bu.,  2150.42  cu.  in.  Irap'l  bu.,  2218.192 cu.  in. 

Weight,         Troy  lb.,  57 GO  gr.  Imperial  lb.,  5760  gr.         Gramme,  15.433159  T.  gr. 

NoTKS. — 1.  An  Imperial  gallon  is  equal  to  1.2  wine  gallons. 

2.  An  old  ale  or  beer  gallon  is  very  nearly  1.221  wine  gallons,  or  1.017  Im- 
perial gallons. 

3.  In  ordinary  computations  2150.4  on.  in.  may  be  taken  as  a  Winchester 
bushel,  and  2218.2  cu.  in.  as  an  Imperial  busheL 


MONEY  AND  CURRENCIES. 

34:3.  Money  is  the  commodity  adopted  to  serve  as  the  uni- 
versal equivalent  or  measure  of  value  of  all  other  commodities, 
and  for  which  individuals  readily  exchange  their  surplus  products 
or  their  services. 

34:4:,    Coin  is  metal  struck,  stamped,  or  pressed  with  a  die,  to 

give   it  a  legal,  fixed  value,  for  the    purpose  of  circulating  as 

money. 

Note.  —  The  coins  of  civilized  nations  consist  of  gold,  silver,  copper,  and 
nickel. 

34ft5.    A  Mint  is  a  place  in  which  the  coin  of  a  country  or 

government  is  manufactured. 

NoTR.  —  In  all  civilized  countries  mints  and  coinage  are  under  the  exclusive 
direction  and  control  of  government. 


188  COMPOUND    NUMBERS. 

S43.     An  Alloy  is  a   metal    compounded   with   another  of 

greater  vahie.     In  coinage,  the  less  vahiable  or  baser  metal  is  not 

reckoned  of  any  value. 

NoTK.— Gold  and  silver,  in  their  pure  state,  are  too  soft  and  flexible  for  coin- 
age; hence  they  are  hardened  by  compounding  them  with  an  alloy  of  baser 
metal,  while  their  color  and  other  valuable  qualities  are  not  materially  impaired. 

34: T*  An  Assayer  is  a  person  who  determines  the  composi- 
tion and  consequent  value  of  alloyed  gold  and  silver. 

The  fineness  of  gold  is  estimated  by  carats,  as  follows :—  * 

Any  mass  or  quantity  of  gold,  either  pure  or  alloyed,  is  divided 
into  24  equal  parts,  and  each  part  is  called  a  carat. 

Fine  gold  is  pure,  and  is  24  carats  fine. 

Alloyed  gold  is  as  many  carats  fine  as  it  contains  parts  in  24  of 
fine  or  pure  gold.  Thus,  gold  20  carats  fine  contains  20  parts  or 
carats  of  fine  gold,  and  4  parts  or  carats  of  alloy. 

348.  An  Ingot  is  a  small  mass  or  bar  of  gold  or  silver,  in- 
tended either  for  coinage  or  exportation.  Ingots  for  exportation 
usually  have  the  assayer's  or  mint  value  stamped  upon  them. 

349.  Bullion  is  uncoined  gold  or  silver. 

3«10.  Bank  Bills  or  Bank  Notes  are  bills  or  notes  issued  by 
a  banking  company,  and  are  payable  to  the  bearer  in  gold  or  silver, 
at  the  bank,  on  demand.  They  are  substitutes  for  coin,  but  are 
not  legal  tender  in  payment  of  debts  or  other  obligations. 

3ol«  Treasury  Notes  are  notes  issued  by  the  General  Govern- 
ment, and  are  payable  to  the  bearer  in  gold  or  silver,  at  the  gene- 
ral treasury,  at  a  specified  time. 

3^S.  Currency  is  coin,  bank  bills,  treasury  notes,  and  other 
substitutes  for  money,  employed  in  trade  and  commerce. 

3«>3.  A  Circulating  Medium  is  the  currency  or  money  of  a 
country  or  government. 

3o4«  A  Decimal  Currency  is  a  currency  whose  denpmina- 
tions  increase  and  decrease  according  to  the  decimal  scale. 

I.  United  States  Money. 

3*]^tS«  The  currency  of  the  United  States  is  decimal  currency, 
and  is  sometimes  called  Federal  Money, 


I 


MONEY  AND  CURRENCIES. 


189 


The  unit  is  the  dollar,  and  all  the  other  denominations  are  either 
divisors  or  multiples  of  this  unit. 

TABLE. 

10  mills  (m.)  make  1  cent, ct. 

10  cents  "      1  dime, d. 

10  dimes  "      1  dollar, $. 

10  dollars  "      1  eagle, E. 

^  UNIT    EQUIVALENTS. 

ct.  m. 

d.  1    =  10 

$  1    =        10    =        100 

E.  1    =      10    =      100    =      1000 

1    ==    10    =    100    =    1000    =    10000 

Scale —«?  uniformly  10. 

NoTKS.  —  1.  Federal  Money  was  adopted  by  Congress  in  1786. 
2.  The  character  $  is  supposed  to  be  a  contraction  of  U.  S.,  (United  States,) 
the  U  being  placed  upon  the  S. 

Coins.  The  gold  coins  are  the  double  eagle,  eagle,  half  eagle, 
quarter  eagle,  three  dollar  piece  and  dollar. 

The  silcer  coins  are  the  half  and  quarter  dollar,  dime  and  half 
dime,  and  three-cent  piece. 

The  nickel  coin  is  the  cent. 

NoTF.s. — 1.  The  foUowinj^  pieces  of  gold  are  in  circulation,  but.are  not  legal 
coin,  viz. :  the  fifty  dollar  piece,  and  the  half  and  quarter  dollar  pieces. 

2.  The  silver  dolhir,  and  the  copper  cent  and  half  cent,  are  no  longer  coined 
for  general  circulation. 

3.  The  mill  is  a  denomination  used  only  in  computations;  it  is  not  a  coin. 

3«IG*  Government  Standard.  By  Act  of  Congress,  January 
18,  1837,  all  gold  and  silver  coins  must  consist  of  9  parts  (.900) 
pure  metal,  and  1  part  (.100)  alloy.  The  alloy  for  gold  must  con- 
sist of  equal  parts  of  silver  and  copper,  and  the  alloy  for  silver  of 
pure  copper. 

The  three-cent  piece  is  3  parts  (f)  silver,  and  1  part  (})  copper. 

The  nickel  cent  is  88  parts  copper  and  12  parts  nickel. 

STATE    CURRENCIES. 

3«>7«  United  States  money  is  reckoned  in  dollars,  dimes,  cents, 
(ind  mills,  one  dollar  being  uniformly  valued  in  all  the  States  at 
100  cents ;  but  in  many  of  the  States  money  is  sometimes  reckoned 
in  dollars,  shillings,  and  pence.  ^ 


190  COMPOUND  NUMBEKS. 

Note.  — At  the  time  of  the  adoption  of  our  decimal  currency  by  Congress,  in 
1786,  the  colonial  eurrtnci/,  or  billn  of  credit,  issued  by  the  coh)nies,  had  depre- 
ciated in  value,  and  this  depreciation,  being  unequal  in  the  different  colonies, 
gave  rise  to  the  different  values  of  the  State  currencies;  this  variation  continues 
wherever  the  denominations  of  shillings  and  pence  are  in  use. 

Georgia  Currency. 
Georgia,  South  Carolina, $1  =  4s.  8d.  =  5Gd. 

Canada  Currency, 

Canada,  Nova  Scotia, %1  =  5s.  =  60d. 

New  England  Currency, 

New  England  States,  Indiana,  Illinois,  ] 

Missouri,  Virginia,  Kentucky,  Tennes-  V §1  =  6s.  ==  72d. 

see,  Mississippi,  Texas, J 

Pennsylvania  Currency. 

New  Jersey,  Pennsylvania,  Delaware,  |  (&i  __  ^g   /^^i   __  oa^i 

Maryland, |  . .  . .  s? 

Neio  York  Currency. 

New  York,  Ohio,  Michigan,  | $1  =  8s.  =  9Gd. 

North  Carolina, j 


II.     Canada  Money. 

338.  The  currency  of  the  Canadian  provinces  is  decimal,  and 

the  table  and  denominations  are  the  same  as  those  of  the  United 

States  money. 

Note.  —  The  decimal  currency  was  adopted  by  the  Canadian  Parliament  in 
1S58,  and  the  Act  took  effect  in  1859.  Previous  to  the  latter  year  the  money 
of  Canada  was  reckoned  in  pounds,  shillings,  and  pence,  the  same  as  in  Eng- 
land. 

Coins.     The  new  Canadian  coins  are  of  silver  and  copper. 

The  silver  coins  are  the  shilling  or  20-cent  piece,  the  dime,  and 

half  dime.  • 

The  copper  coin  is  the  cent. 

Note.  —  The  20-cent  piece  represents  the  value  of  the  shilling  of  tho  old 
Canada  Currency. 

339.  Government  Standard.    The  silver  coins  consist  of  925 

parts  (.925)  pure  silver  and  75  parts  (.075)  copper.    That  is,  they 

are  .925  fine. 

NoTR.  —  The  value  of  the  20-cent  piece  in  United  States  money  is  18§  cents, 
of  the  dime  9J  cents,  and  of  the  half  dime  4^  cents. 


MONEY  AND  CURRENCIES. 


191 


360.  English  or  Sterling  Money  is  the  currency  of  Great 
Britain.  i^ 

The  unit  is  the  pound  sterling,  and  all  the  other  denominations 
are  divisors  of  this  unit. 

TABLE. 

4  farthings  (far.  or  qr.)  make  1  penny, d. 

12  pence  "       1  shilling, s. 

20  shillings  **       1  pound  or  sovereign  ,,  £  or  sov. 

UNIT    EQUIVALENTS. 

d.  far. 

«•  1=4 

£,  or  sov.    1  =     12  =^    48 

1  =r  20  =  240  =  960 
Scale  —  ascending,  4,  12,  20 ;  descending,  20,  12,  4. 

Notes.  —  1.  Farthings  are  generally  expressed  as  fractions  of  a'penny;  thus, 

1  far.,  sometimes  called  1  quarter,  (qr.)=--id.;  3  fnr.  =|d. 

2.  The  old/,  the  original  abbreviation  fur  shillings,  was  formerly  written  be- 
tween shillings  and  pence,  and  d,  the  abbreviation  for  pence,  was  omitted.  Thus 
2s.  6d.  was  written  2/6.  A  straight  line  is  now  used  in  place  of  the/,  and  shil- 
lings are  written  on  the  left  of  it  and  pence  on  the  right.  Thus,  2/6,  10/3, 
«tc. 

Coins.  The  gold  coins  are  the  sovereign  (=  £1)  and  the  half 
sovereign,  (=  10s.) 

The  silvei'  coins  are  the  crown  (=  5s.),  the  half  crown  (=  2s. 
6d.),  the  shilling,  and  the  6  penny  piece. 

The  copper  coins  are  the  penny,  half  penny,  and  farthing. 

Note. — The  guinea  (=  21s.)  and  the  half  guinea  (=  10s.  6d.  sterling)  are  old 
gold  coins,  that  are  still  in  circulation,  but  are  no  longer  coined. 

301*  Government  Standard.  The  standard  fineness  of  Eng- 
lish gold  coin  is  11  parts  pure  gold  and  1  part  alloy;  that  is,  it  is 
22   carats  fine.     The  standard  fineness  of  silver  coin  is  11  oz. 

2  pwt.  (=  11.1  oz.)  pure  silver  to  18  pwt.  (=  .9  oz.)  alloy.  Hence 
the  silver  coins  are  11  oz.  2  pwt.  fine;  that  is,  11  oz.  2  pwt.  pure 
silver  in  1  lb.  standard  silver. 

This  standard  is  37  parts  (|J  =  .925)  pure  silver  and  3  parts 
(?%  =  .075)  copper. 

Note.  —  A  pound  of  English  standard  gold  is  equal  in  value  to  14.2878  lb.  =» 
14  lb.  3  oz.  9  pwt.  1.727  gr.  of  silver. 


192  COMPOUND  NUMBERS. 

TV.     French  Money. 
30^.    The  currency  of  France  is  decimal  currency. 
The  unit  is  the  franc,  of  which  the  other  denominations  are 
divisors. 

TABLE. 

10  millimes  make  1  centime. 
100  centimes     "      1  franc. 
Scale  —  ascending,  10,  100;  descending,  100,10. 

Coins.     The  gold  coin  is  the  20-franc  piece,  or  Louis. 

The  silver  coins  are  the  franc  and  the  demi  franc. 

Note. — In  France  accounts  are  kept  in  francs  and  decimes.    A  franc  is  equal 
to  18.6  cents  U.  S.  money. 

363.  COMPARATIVE  TABLE  OF  MONEYS. 


English.                                         U.  S. 

French. 

U.S. 

Iqr.                       =  $  .004 1 J 

Id.                =    -02;^ 

Is.                       =      .242 
4s.  Id.  2xVrqr.  =    LOO 
£1                      =   4.84 

1  millime  = 
1  centime  = 
1  franc       = 

$  .000186 
.00186 
.186 

REDUCTION. 

364.  Reduction  is  the  process  of  changing  a  number  from 
one  denomination  to  another  without  altering  its  value. 

Reduction  is  of  two  kinds,  Descending  and  Ascending. 

363.  Reduction  Descending  is  changing  a  number  of  one 
denomination  to  another  denomination  of  less  unit  value;  thus, 
$1  =  10  dimes  =  100  cents  =  1000  mills. 

306.  Reduction  Ascending  is  changing  a  number  of  one 
denomination  to  another  denomination  of  greater  unit  value  ;  thus, 
1000  mills  =  100  cents  =  10  dimes  =  $1. 

REDUCTION    DESCENDING. 
CASE   I. 

367.  To  reduce  a  compound  number  to  lower  de- 
nominations. 

1.  Reduce  3  mi.  1  fur.  17  rd.  2  yd.  1  ft.  8.  in.  to  inches. 


REDUCTION. 


193 


OPERATION. 

3  mi.  1  fur.  17  rd.  2  yd.  1  ft.  8  in. 


25 
40 


fur. 


1017  rd. 
5} 

5087 
508} 

5595}  yd. 
3 

16787}  ft. 
12 

201458  in. 

Analysis.  Since  in 
1  mi.  there  are  8  fur., 
in  3  miles  there  are 
3X8  fur.  ==  24  fur., 
and  the  1  fur.  in  the 
given  number,  added, 
makes  25  fur.  in  3  mi. 
1  fur.  Since  in  1  fur. 
there  are  40  rd.,  in  25 
fur.  there  are  25  X  40 
rd.  =  1000  rd.,  and 
the  17  rd.  in  the  given 
number  added,  makes 
1017  rd.  in  3  mi.  1  fur. 
17  rd.  Since  in  1  rd. 
there  are  5}  yd.,  in 
1017  rd.  there  are 
1017  X  5}  yd.  =  5503}  yd.,  which  plus  the  2  yd.  in  the  given  number 
=  5595}  yd.  in  3  mi.  1  fur.  17  rd.  2  yd.  Since  in  1  yd.  there  are 
3  ft.,  in  5595}  yd.  there  are  5595}  X  3  ft.  =  16786}  ft.,  which  plus 
the  1  ft.  in  the  given  number  =  16787}  ft.  in  3  mi.  1  fur.  17  rd.  2  yd. 
1  ft.  And  since  in  1  ft.  there  are  12  in.,  in  16787}  ft.  there  are 
16787}  X  12  in.  ==  201450  in.,  which  plus  the  8  in.  in  the  given  num- 
ber =  201458  in.  in  the  given  compound  number.  On  examining  the 
operation,  we  find  tbj.t  we  have  successively  multiplied  by  the  numbers 
in  the  descending  scale  of  linear  measure  from  miles  to  inches,  inclu- 
sive. But,  as  either  factor  may  be  used  as  a  multiplicand,  (82, 1),  we 
may  consider  the  numbers  in  the  descending  scale  as  multipliers. 
Hence  the  following 

Rule.  I.  Multiply  the  JiigJiest  denomination  of  tJie  given 
compound  number  hy  that  number  of  the  scale  which  will  reduce  it 
to  the  next  lower  denomination,  and  add  to  the  product  the  given 
numbery  if  any,  of  that  lower  denomination, 

II.  Proceed  in  the  same  manner  with  the  results  obtained  in  each 
lower  denominationj  until  the  reduction  is  brought  to  the  denomina- 
tion required. 

EXAMPLES    FOR   PRACTICE. 

1.  In  16  lb.  10  oz.  18  pwt.  5  gr.,  how  many  grains? 
17 


194  COMPOUND  NUMBERS. 

2.  In  £133  6  s.  8d.,  how  many  farthings?     Ans.  128,000. 

3.  Change  100  mi.  to  inches.  Ans.  6336000  in. 

4.  How  many  rods  of  fence  will  inclose  a  farm  IJ  miles 
square  ?  Ans.  1920  rd. 

5.  The  grey  limestone  of  Central  New  York  weighs  175  lbs. 
to  the  cubic  foot;  what  is  the  weight  of  a  block  8  ft.  long  and 
1  yd.  square  ?  Ans.  6  T.  6  cwt. 

6.  What  will  be  the  cost  of  1  hhd.  of  molasses  at  $.28  per  gal.  ? 

7.  A  man  wishes  to  ship  1548  bu.  1  pk.  of  potatoes  in  barrels 
containing  2  bu.  3  pk.  each ;  how  many  barrels  must  he  obtain  ? 

8.  A  grocer  bought  10  bu.  of  chestnuts  at  $3.75  a  bushel,  and 
retailed  them  at  §.06i  a  pint;  how  much  was  his  whole  gain  ? 

9.  Eeduce  90^  17'  40''  to  seconds.  Ans.  325060". 

10.  In  the  18th  century  how  many  days  ?      Ans.  36524  da. 

11.  At  6}  cts.  each,  what  will  be  the  cost  of  a  great-gross  of 
writing  books  ?  Ans.  $108. 

12.  How  large  an  edition  of  an  octavo  book  can  be  printed 
from  4  bales  4  bundles  1  ream  10  quires  of  paper,  allowing  8 
sheets  to  the  volume  ?  A)is.  2970  vol. 

13.  Suppose  your  age  to  be  18  yr.  24  da.  ;  how  many  minutes 
old  are  you,  allowing  4  leap  years  to  have  occurred  in  that  time  ? 

14.  How  many  pence  in  481  sovereigns?      Am.  115,440  d. 

15.  Reduce  $7J  to  mills.  Ani.  7375  mills. 

16.  In  3  P.  of  Sherry  wine,  how  many  qt.  ?      Ans.  1560  qt. 

17.  Reduce  37  Eng.  ells  1  qr.  to  yd.         Ans.  46  yd.  2  qr. 

18.  In  £6  10s.  lOd.  how  many  dollars  U.  S.  currency  ? 

19.  Reduce  6,0.  14fg  Sf^  45iri  ^^  minims. 

20.  Reduce  1  T.  1  P.  1  hhd.  to  Imperial  gallons. 

Ans.  367  J  Imperial  gal. 

21.  How  many  dollars  Canada  currency  are  equal  to  £126  12s. 
6d.?  Ans.  $506i. 

22.  How  many  pint,  quart,  and  two-quart  bottles,  of  each  an 
equal  number,  may  be  filled  from  a  hogshead  of  wine  ? 

Ans.   72. 

23.  IIow  many  steps  of  2  ft.  9  in.  each,  will  a  man  take,  in 
walking  from  Erie  to  Cleveland,  the  distance  being  95  mi.? 


PvEDUCTION. 


195 


BC 


24.  A  grocer  bought  12  bbl.  of  cider  at  $11  a  barrel,  and  after 
converting  it  into  vinegar,  he  retailed  it  at  6  cents  a  quart  ]  how 
much  was  his  whole  gain  ?  Ans.  $69.72. 

25.  In  75  A.  4  sq.  ch.  18  P.  118  sq.  1.  how  many  square  links? 

26.  How  many  inches  high  is  a  horse  that  measures  16  hands  ? 

27.  If  a  vessel  sail  150  leagues  in  a  day,  how  many  statute 
miles  does  she  sail  ?  Ans.  517.5. 

28.  If  14  A.  be  sold  from  a  field  containing  50  A.,  how  many 
gquare  rods  will  the  remainder  contain  ?         Ans.  5,760  sq.  rd. 

29.  A  man  returning  from  Pike's  Peak  has  36  lb.  8  oz.  of 
ure  gold ;  what  is  its  value  at  $1.04i  per  pwt.  ?    Ans.  ^9169.60. 

30.  A  person  having  8  hhd.  of  tobacco,  each  weighing  9  cwt. 
42  lb.,  wishes  to  put  it  into  boxes  containing  48  lb.  each ;  how 
many  boxes  must  he  obtain  ?  Ans.  157. 

31.  A  merchant  bought  12  bbl.  of  salt  at  $li  a  barrel,  and  re- 
tailed it  at  f  of  a  cent  a  pound;  how  much  was  his  whole  gain? 

32.  A  physician  bought  lib  lOg  of  quinine  at  §2.25  an  ounce, 
and  dealt  it  out  in  doses  of  10  gr.  at  $.12 J  each;  how  much  more 
than  cost  did  he  receive?  Ans.  $82.50. 


CASE   II. 

368.   To  reduce  a  denominate  fraction  from  a  greater 
to  a  less  unit. 

1.  Reduce  -^^  of  a  gallon  to  the  fraction  of  a  gill. 

AXALYSIS. 


OPERATION. 

4i  gal.  X  I  X  f  X  I 
Or, 
11 


=  TT  gl- 


V^ 


To  re- 
duce gallons  to  gills, 
we  multiply  succes- 
sively by  4,  2,  and  4, 
the  numbers  in  the  de- 
scending scale.  And 
since  the  given  num- 
ber is  a  fraction,  we 
indicate  the  process, 
as  in  multiplication  of 
fractions,  after  which  we  perform  the  indicated  operations,  and  ob- 
tain j\,  the  answer.     Hence, 


11 


1 
4 
2 

8  =  TT  g^-^  ^■^«- 


196  COMPOUND  NUMBERS. 

KuLE.     Multiply  the  fraction  of  the  higher  denomination  hy  the 
•^.    numhei's  in  the  desc ending  scale  successively ^  between  the  given  and 
the  required  denomination. 
Note.  —  Cancellation  may  be  applied  wherever  practicable. 

EXAMPLES    FOR   PRACTICE. 

1.  Eeduce  ^^^  of  a  lb.  Troy  to  the  fraction  of  a  pennyweight. 

Ans.   I  pwt. 

2.  Eeduce  ^^^  of  a  hhd.  to  the  fraction  of  a  pint. 

3.  Reduce  ^ jy^  of  a  mile  to  the  fraction  of  a  yard. . 

Ans.  I  yd. 

4.  Eeduce  -g|^  of  a  gallon  to  the  fraction  of  a  gill. 

5.  What  part  of  a  dram  is  -^^^-q  of  |  of  |  of  -^j  of  S|  pounds 
avoirdupois  weight ?  Ans,  :^2T5  ^^* 

6.  Eeduce  yg^^^j  of  a  dollar  to  the  fraction  of  a  cent. 

7.  Eeduce  ^K  of  a  rod  to  the  fraction  of  a  link.     Ans,  |  1. 

8.  Eeduce  -^K  of  a  scruple  to  the  fraction  of  a  grain. 

9.  What  fraction  of  a  yard  is  ^  of  ■j\  of  a  rod  ? 

10.  ^^g  of  -a  week  is  |  of  how  many  days?  Ans.  8|  da. 

11.  What  fraction  of  a  square  rod  is  j^^g^  of  4|  times  y^^  of  an 
acre  ?  Ans,  j\  sq.  rd. 

CASE  III.  jl 

369.  To  reduce  a  denominate  fraction  to  integers 

of  lower  denominations.  ,, 

1.  What  is  the  value  of  |  of  a  bushel? 

OPERATION.  Analysis.     |  bu.  =  |  of 

I  bu.  X  4  =   f  pk.  =  If  pk.  4  pk.,  or  1§  pk.;  }  pk.  =  ^ 

I  pk.  X  8  =  \4  qt.  ==  4|  qt.  of  8  qt.  =  4|  qt. ;  and  J  qt. 

I  qt.   X  2  =   i   pt.  =  If  pt.  =  I  of  2  pt.  =  15  pt.    The 

1  pk.  4  qt.  If  pt,  Ans.  "^^*^'   1  P^-'  ^  ^*"  1  P*" 

with   the   last  denominato 

fraction,  |  pt.,  form  the  answer.     Hence, 

EuLE.     I.  Multiply  the  fraction  hy  that  number  in  the  scale 

which  will  reduce  it  to  the  next   lower  denomination^  and  if  the 

result  be   an  improper  fraction^  reduce  it  to  a  whole  or  mixed 

number. 


EEDUCTIOK. 


19T 


II.  Proceed  with  the  fractional  i-)artj  if  any^  as  before,  until 
reduced  to  the  denominations  required.  ^ 

III.  The  units  of  the  several  denominations j  arranged  in  their 
ordery  ivill  he  the  required  residt. 


i 


EXAMPLES    FOR   PRACTICE. 

1.  Keduce  -^j^  of  a  yard  to  integers  of  lower  denominations. 

Ans.  2  ft.  8^  in. 

o 

2.  Reduce  |  of  a  month  to  lower  denominations. 

3.  Reduce  g^J  of  a  short  ton  to  lower  denominations. 

4.  What  is  the  value  of  |  of  a  long  ton  ? 

Ans.  11  cwt.  12  lb.  7^  oz. 


? 


5.  What  is  the  value  of  |  of  2^  pounds  apothecaries'  weight  ? 

6.  What  is  the  value  of  -^^  of  an  acre  ? 
Ans.  2  R.  6  P.  4  sq.  yd.  5  sq.  ft.  127/^  sq.  in. 

7.  Reduce  |  of  a  mile  to  integers  of  lower  denominations. 

8.  What  is  the  value  of  4  of  a  great  gross  ? 

Ans.  6  gross  10  doz.  8|. 

9.  What  is  the  value  in  geographic  miles  of  -^^  of  a  great 
circle?  Ans.  12150  mi. 

10.  What  is  the  value  of  |  of  3|  cords  of  wood  ? 

Ans.  2  Cd.  5  cd.  ft.  9|  cu.  ft. 

11.  The  distance  from  Buffalo  to  Cincinnati  is  438  miles;  hav- 
ing traveled  |  of  this  distance,  how  far  have  I  yet  to  travel  ? 

Ans,  262  mi.  G  fur.  16  rd. 

12.  What  is  the  value  of  ||  f §  ?  Ans.  8  f^  35  rt^. 


13.  What  is  the  value  of  -|  of  a  sign  ? 


Ans.  12°  5r  25^". 


14.    A  man  having  a  hogshead  of  wine,    sold  -^^  of  it;  how 
much  remained  ?  Ans.  S3  gal.  3  qt.  1  pt.  l-j^^  gi. 


CASE   IV. 


S7®.  To  reduce  a  denominate  decimal  to  integers 
of  lower  denominations. 

1.  Reduce  .125  of  a  barrel  to  integers  of  lower  denominations. 
17* 


198  COMPOUND  NUMBERS. 

OPERATION.  Analysis.      We   first  multiply 

I  125  the  given  decimal,  .125  of  a  barrel, 

31.5  by  31.5  (=  31J)   to  reduce  it  to 

o  r%.-,--       1  callons,  and  obtain  3.9375  g-allons. 

8.93/0  gal.  °    .^,.       ^1     Q       11  1 

1  ^  Omitting  the  o  gallons,  we  mul- 

tiply  the  decimal,  .9375   gal.,  by 

3.7500  qt.  4  ^o  reduce  it  to  quarts,  and  obtain 

_f  3.75    quarts.     We   next  multiply 

1.50  pt.  the  decimal  part  of  this  result  by 

4  2,  to  reduce  it  to  pints,  and  obtain 

2  0  0-1  1-^  pints.     And  the  decimal  part 

3  gal.  3  qt.  iVt.  2  gi.,  Ans.     ''^  ^^'^  "-f^"'*  ^^^^  '""'*'P'y  ^'^.^  ^^ 

reduce   it  to   gills,    and  obtain  2 

gills.     The  integers  of  the  several  denominations,  arranged  in  their 

order,  form  the  answer.     Hence, 

E-ULE.  I.  3Iultip7//  the  given  denominate  decimal  hy  that  num- 
her  in  the  descending  scale  ichich  will  reduce  it  to  the  next  loiver 
denominatiouy  and  point  off  the  result  as  in  midtipUcation  of 
decimals, 

II.  Proceed  icifh  the  decimal  part  of  the  product  in  the  same 
manner  until  reduced  to  the  required  denominations.  The  integers 
at  the  left  will  he  the  answer  required. 

EXAMPLES    FOR   rRACTICE. 

1.  What  is  the  value  of  .645  of  a  day? 

Ans.  15  h.  28  min.  48  sec. 

2.  What  is  the  value  of  .765  of  a  pound  Troy  ? 

3.  What  is  the  value  of  .6625  of  a  mile? 

4.  What  is  the  value  of  .8469  of  a  degree  ? 

Ans.  50'  48.84''. 

5.  What  is  the  value  of  .875  of  a  hhd.  ? 

6.  What  is  the  value  of  £.85251  ? 

Ans.  17  s.  2.4  +  far. 

7.  What  is  the  value  of  .715°  ?  Ans.  42'  54". 

8.  What  is  the  value  of  7.88125  acres? 

Ans.  7  A.  3  E.  21  P. 

9.  What  is  the  value  of  .625  of  a  f^ithom?  Ans.  3|  ft. 


REDUCTION. 


199 


10.  What  is  the  value  of  .375625  of  a  barrel  of  pork? 

11.  What  is  the  value  of  .1150390625  Cong.  ?  ^ 

Ans.  Ufg  5f3  48  T?i.     ^' 

12.  What  is  the  value  of  .61  of  a  tun  of  wine  ? 

A71S.  1  P.  27  gal.  2  qt.  1  pt.  3.04  gi. 

REDUCTION    ASCENDING. 
CASE    I. 

371.    To  reduce  a  denominate  number  to  a  com- 
pound number  of  higher  denominations. 

1.  Reduce  157540  minutes  to  weeks. 

Analysis.       Dividing 

the  given  number  of 
minutes  by  60,  because 
there  are  ^'j  as  many 
hours  as  minutes,  and  wo 
obtain  2645  h.  plus  a  re- 
mainder of  40  min.  AYe 
next  divide  the  2645  h. 
by  24,  because  there  are 
-^\  as  many  days  as  hours,  and  we  find  that  2645  h.  =  109  da.  plus  a 
remainder  of  9  h.  Lastly  we  divide  the  109  da.  by  7,  because  there 
are  \  as  many  weeks  as  days,  and  we  find  that  109  da.  =  15  wk.  plus 
a  remainder  of  4  da.  The  last  quotient  and  the  several  remainders 
annexed  in  the  order  of  the  succeeding  denominations,  form  the 
answer. 

2.  Eeduce  201458  inches  to  miles. 

OPERATION. 

12  )  201458  in. 


OPERATION. 

60)157540  min. 

24  )  2625  h.  +  40  min. 

7)109  da.  -f  9  h. 

15  wk.  4-  4  da. 
15  wk.  4  da.  9  h.  40  min.,  Ans. 


3)16788  ft.  2  in. 
5i  or  5.5)5596  yd. 

40  )    1017  rd.  2  yd.  1  ft.  6  in. 
8)  25  fur.  17  rd. 
3  mi.  1  fur. 
3  mi.  1  fur.  17  rd.  2  yd.  1  ft.  8  in.,  Ans. 


Analysis.  We 
divide  successively 
by  the  numbers  in 
the  ascending  scale 
of  linear  measure, 
in  the  same  manner 
as  in  the  last  pre- 
ceding operation. 
But,  in  dividing  the 
5596  yd.  by  5J  or 
5.5,  we  have  a  ro- 


200  COMPOUND  NUMBERS. 

mainder  of  2J  yd.,  and  this  reduced  to  its  equivalent  compound  num- 
ber, (369)  =  2  yd.  1  ft.  G  in.  In  forming  our  final  result,  the  6  in. 
of  this  number  are  added  to  the  first  remainder,  2  in.,  making  the  8  in. 
as  given  in  the  answer.  From  these  examples  and  analyses  we  deduce 
the  following 

Rule.  I.  Divide  the  given  concrete  or  denominate  number  hy 
that  number  of  the  ascending  scale  which  will  reduce  it  to  the  next 
higher  denomination. 

II.  Divide  the  quotient  hy  the  next  higher  number  in  the  scale ; 

and  so  proceed  to  the  highest  denomination  required.      The  last 

quotienty  with  the  several  remainders  annexed  in  a  reversed  order ^ 

will  be  the  answer. 

Note.  —  The  several  corresponding  cases  in  reduction  descending  and  reduo- 
tion  ascending,  being  opposites,  mutually  prove  each  other. 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  1913551  drams  to  tons. 

2.  In  97920  gr.  of  medicine  how  many  lb.  ?        Ans.  17  lb. 
8.  Reduce  1000000  in.  to  mi. 

4.  How  many  acres  in  a  field  120  rd.  long  and  56  rd.  wide  ? 
6.  In  a  pile  of  wood  60  ft.  long,  15  ft.  wide,  and  10  ft.  high, 
how  many  cords  ?  Ans.  70  Cd.  2  cd.  ft.  8  cu.  ft 

6.  How  many  fathoms  deep  is  a  pond  that  measures  28  ft.  6 
in.  ?  Ans.  4f  fath. 

7.  In  30876  gi.  how  many  hhd.  ? 

8.  How  many  bushels  of  corn  in  27072  qt.  ?     Ans.  846  bu. 

9.  At  2  cts.  a  gill,  how  much  alcohol  may  be  bought  for  $2.54? 

10.  In  1234567  far.  how  many  £?  Ans.  £1286  IJ  d. 

11.  Reduce  2468  pence  to  half  crowns. 

12.  In  $88.35  how  many  francs?  Ans.  475. 

13.  In  622080  cu.  in.  how  many  tons  of  round  timber  ? 

14.  In  84621  Tli  how  many  Cong.  ? 

15.  If  135  million  Gillott  steel  pens  are  manufactured  yearly^ 
how  many  great-gross  will  they  make  ?  A7is.  78125. 

16.  Reduce  1020300''  to  S.  Ans.  9  S.  13°  25' 

17.  In  411405  sec.  how  many  da.  ? 


REDUCTION.  201 

18.  During  a  storm  at  sea,  a  ship  changed  her  latitude  412 
geographic  miles  ;  how  many  degrees  and  minutes  did  she  change  ? 

Ans.  G^  52'. 

10.  If  a  man  travel  at  the  rate  of  a  minute  of  distance  in  20 
minutes  of  time,  how  much  time  would  he  require  to  trtivel  round 
the  earth  ?  Ans.  SCO  days. 

20.  In  120  gross  how  many  score  ?  Ans.  864. 

21.  How  many  miles  in  the  semi-circumference  of  the  earth  ? 

22.  How  much  time  will  a  person  gain  in  86  yr.  by  rising  45 
min.  earlier,  and  retiring  25  min.  later,  every  day,  allowing  for  9 
leap  years?  Ans.  689  da.  4  h.  30  min. 

28.  A  grocer  bought  20  gal.  of  milk  by  beer  measure,  and  sold 
it  by  wine  measure ;  how  many  quarts  did  he  gain  ?     Ans.  17f  4. 

24.  How  many  bushels  of  oats  in  Connecticut  are  equivalent  to 
1500  bushels  in  Iowa?  Ans.  1875  bu. 

25.  Reduce  120  leagues  to  statute  miles.  Ans.  414  mi. 

26.  In  1  bbl.  1  gal.  2  qt.  wine  measure,  how  many  beer  gal- 
lons ?  Ans.  27-5^. 

27.  Reduce  150  U.  S.  bushels  to  Imperial  bushels. 

Ans.  145.415  +  Imp'l.  bu. 

28.  How  many  squares  in  a  floor  68  ft.  8  in.  long,  and  88  ft. 
wide?  Ans.  22^. 

29.  How  many  cubic  inches  in  a  solid  4  ft.  long  8  ft.  wide,  and 
1  ft.  6  in.  thick  ? 

30.  How  many  acres  in  a  field  120  rd.  long  and  56  rd.  wide  ? 

31.  Change  856  dr.  apothecaries  weight,  to  Troy  weight. 

82.  A  coal  dealer  bought  175  tons  of  coal  at  S8.75  by  the  long 
ton,  and  sold  it  at  §4.50  by  the  short  ton ;  how  much  was  his 
whole  gain  ?  Aiis.  $225.75. 

38.  How  many  acres  of  land  can  be  purchased  in  the  city  of 
New  York  for  $78750,  at  $1.25  a  square  foot? 

Ans.   1  A.  56  P.  104  sq.  ft. 

84.  An  Ohio  farmer  sold  a  load  of  corn  weighing  2402  lb.,  and  a 
load  of  wheat  weighing  2175  lb. ;  for  the  corn  he  received  S.60  a 
bushel,  and  for  the  wheat  $1.20  a  bushel;  how  much  did  he  re* 
ceive  for  both  loads  ?  Ans.  $70.20. 


202 


COMPOUND  NUMBERS. 


The  following  examples  are  given  to  illustrate  a  short  and  prac- 
tical method  of  reducing  currencies. 

85.  What  will  be  the  cost  of  54  bu.  of  corn  at  5s.  a  bushel, 
New  England  currency  ? 


OPERATION. 


54  X  5  =    270s. 
270s.  -V-  6  =  $45 


OPERATION. 

3 

m 

Or,  i 

100          ^ 

$300 

6 

25 

2 

$300 

Or,  9         Analysis.    Since  1  bu.  costs 

^^       5s.,  54  bu.  cost  54x  5s.  =  270s.; 

--  and  since  6s.  make  $1  N.  E. 

r-p      currency,  270  —-  G  =  $45,  Ans. 
^45 

36.  What  will  270  bu.  of  wheat  cost,  @  8s.  4d.  Penn.  currency  ? 

Analysis.  Multiply  the 
quantity  by  the  price  in  Penn. 
currency,  and  divide  the  cost 
by  the  value  of  $1  in  the  same 
currency;  or  reduce  the  shil- 
lings and  pence  to  a  fraction 
of  a  shilling,  before  multiply- 
ing and  dividing. 

37.  Bought  5  hhd.  of  rum  at  the  rate  of  2s.  4d.  a  quart,  Geor- 
gia currency ;  how  much  was  the  whole  cost  ? 

OPERATION. 

5 
63 
Or,  2 

i  t 

$630 

88.  Sold  120  barrels  of  apples,  each  containing  2  bu.  2  pk.,  at 
4s.  7d.  a  bushel,  and  received  pay  in  cloth  at  10s.  5d.  a  yard ;  how 
many  yards  of  cloth  did  I  receive  ? 

Analysis.  The  operation  in  this  example  is 
similar  to  the  preceding  examples,  except  that  we 
divide  the  cost  of  the  apples  by  the  ^rice  of  a  unit 
of  the  article  received  in  payment,  reduced  to  units 
of  the  same  denomination  as  the  price  of  a  unit  of 
the  article  sold.  The  result  will  be  the  same  in 
$lo2  whatever  currency. 


5 
63 

2 

i 


$630 


OPERATION. 

12 

11 


Analysis.  In  this  ex- 
ample we  first  reduce  5 
hhd.  to  quarts  by  multiply- 
ing by  63  and  4,  and  then 
proceed  as  in  the  preceding 
examples. 


m 


KEDUCTION.  203 

39.  What  cost  75  yards  of  fianncl  at  3s.  Cd.  per  yard,  New 
England  currency  ?  Ajis.  813.75. 

40.  A  man  in  Philadelphia  worked  5  weeks  at  Gs.  4d.  a  day; 
how  much  did  his  wages  amount  to  ?  Ans.   $25.33}. 

41.  A  farmer  exchanged  2  bushels  of  beans  worth  10s.  6d.  per 
bushel,  for  two  kinds  of  sugar,  the  one  at  lOd.  and  the  other  at 
lid.  per  pound,  taking  the  same  quantity  of  each  kind;  how 
many  pounds  of  sugar  did  he  receive  ?  Ans.  24  lb. 

42.  If  corn  be  rated  at  6s.  lOd.  per  bushel  in  Vermont,  at  what 
price  in  the  currency  of  New  Jersey  must  it  be  sold,  in  order  to 
gain  §7.50  on  54  bushels? 

CASE    IT. 

37@.  To  reduce  a  denominate  fraction  from  a  less  to 
a  greater  unit. 

1.  Reduce  -fj  of  a  gill  to  the  fraction  of  a  gallon. 

OPERATION.  Analysis.     To   re- 

t\  gi-  X  I  X  ^  X  i  =  4^  gal.  ^^c«,  g'^lls  to  gallons, 

we  divide  successively 
^^>  by   4,    2,    and  4,  the 

11       ^  numbers    in   the    as- 

4  cending   scale.      And 

since  the  given  num- 

ber  is  a  fraction,  we 

44       1  =  -^^j  Ans.  indicate  the  process, 

as  in  division  of  frac- 
tions, after  which  we  perform  the  indicated  operations,  and  obtain 
^^,  the  answer.     Hence, 

Rule.  Divide  the  fraction  of  the  lower  denomination  hi/  the 
numbers  in  the  ascending  scale  successively/ ,  between  the  given  and 
the  required  denomination. 

Note. — The  operation  may  frequently  be  shortened  by  cancellation. 

EXAMPLES    FOR   PRACTICE. 


1.  Reduce  f  of  a  shilling  to  the  fraction  of  a  pound. 

Ans.  £ 


^u- 


204  COMPOUND  NUMBERS. 

2.  Keduce  ^  of  a  pennyweight  to  the  fraction  of  a  pound 
Troy.  Ans.  ^J^  lb. 

3.  What  part  of  a  ton  is  |  of  a  pound  avoirdupois  weight  ? 

4.  What  fraction  of  an  hour  is  |  of  20  seconds  ? 

6.  What  is  the  fractional  difference  between   g|^  of  a  hhd. 
and  I  of  a  pt.  ?  Ans.   35^77^  hhd. 

6.  ^1^  of  I  of  f  of  a  pint  is  what  fraction  of  2  pecks  ? 

Ans.  I 

7.  Reduce  |  of  ^   of  /$   of  a  cord  foot  to  the  fraction  of  a 
cord.  Ans.  -^g  Cd. 

8.  What  part  of  an  acre  is  ^^^  of  jly  of  9^  square  rods  ? 

9.  I  of  5^  furlongs  is  ^  of  J^  of  how  many  miles? 

Ans.  12|  mi. 

10.  A  block  of  granite  containing  |  of  |  of  20^  cubic  feet,  is 
what  fraction  of  a  perch  ?  Ans.  ^|  Pch. 

11.  What  part  of  a  cord  of  wood  is  a  pile  7^  ft.  long,  2  ft.  high, 
and  3|  feet  wide  ?  Ans.  Iff  Cd. 

12.  Reduce  |  of  an  inch  to  the  fraction  of  an  Ell  English. 


CASE   III. 

S73.  To  reduce  a  compound  number  to  a  fraction  of 
a  higher  denomination. 

1.  Reduce  2  oz.  12  pwt.  12  gr.  to  the  fraction  of  a  pound  Troy. 

OPERATION.  Analysis.      To   find 

2  oz.  12  pwt.  12  gr.  =  1260  gr.  what  part  one  compound 

1  lb.  Troy  =  5760  gr.  number   is   of  another, 

Jf  §g  lb.  =  ^5  lb.,  Ans.        they  must  be  like  num- 
bers and  reduced  to  the 
same  denomination.     In  2  oz.  12  pwt.  12  gr.  there  are  1260  gr.,  and 
in  1  lb.  there  are  5760  gr.     Therefore  1  gr.  is  3^'g^^  lb.,  and  1260  gr. 
are  f^|^  lb.  ==  3^  lb->  the  answer.     Hence, 

Rule.  Reduce  the  given  number  to  its  lowest  denoniination  for 
the  numerator  J  and  a  unit  of  the  required  denomination  to  the  same 
denomination  for  the  denominator  of  the  required  fraction. 

Note. — If  the  given  number  contain  a  fraction,  the  denominator  of  this  frac- 
tion must  be  regarded  as  the  lowest  denomination. 


REDUCTION.  _,  205 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  2  R.  20  P.  to  the  fraction  of  an  acre. 

Ans.  I  A. 

2.  What  part  of  a  mile  is  6  fur.  26  rd.  3  yd.  2  ft.  ? 

3.  What  part  of  a  £  is  18s.  5d.  2/^  far.  ?  Ans.  £if . 

4.  What  part  of  21  lb.  Apothecaries'  weight  is  7g  73  29  14 

5.  What  part  of  3  weeks  is  4*  da.  16  h.  30  min.  ? 

6.  Reduce  1|  pecks  to  the  fraction  of  a  bushel. 

7.  From  a  hogshead  of  molasses  28  gal.  2  qt.  were  drawn ; 
what  part  of  the  whole  remained  in  the  hogshead?       Ans.  ||. 

8.  Reduce  4  bundles  6  quires  16  sheets  of  paper  to  the  frac- 
tion of  a  bale.  Ans.   |  of  a  bale. 

9.  What  part  of  54  cords  of  wood  is  4800  cubic  feet  ? 

10.  What  is  the  value  of  (^  of  a  dollar  ?  Ans,  86.30. 

11.  Reduce  30.  of  3  If  3  80  nj,  to  the  fraction  of  a  Cong. 

12.  What  part  of  a  ton  of  hewn  timber  is  06  cu  ft.  864  cu.  in.  ? 

CASE   IV. 

874.  To  reduce  a  compound  number  to  a  decimal  of 
a  higher  denomination. 

1.  Reduce  3  cd.  ft.  8  cu.  ft.  to  the  decimal  of  a  cord. 

OPERATION.  Analysis.     Were- 

16     8.0  cu.  ft.  duce  the  8  cu.  ft.  to 

the  decimal  of  a  cd. 


3.5000  cd.  ft. 


ft.,  by  annexing  a  ci- 

.4375  Cd.,  Ans,  pher,    and    dividing 

Or,  by  16,  the  number  of 

3  cd.  ft.  8  cu.  ft.  =  56  cu.  ft.  cu.  ft.  in  1  cd.  ft.,  an- 

1  Cd.       =  128  cu.  ft.  nexing    the    decimal 

^s_fig  Cd.  =  ^5  Cd.  =  .4375  Cd.,  Ans.         quotient  to  the  3  cd. 

ft.     We   now  reduce 
the  3.  5  cd.  ft.  to  Cd.  or  a  decimal  of  a  Cd.,  by  dividing  by  8,  the 
number  of  cd.  ft.  in  1  Cd.,  and  we  have  .4375  Cd.,  the  answer. 
Or,  we  may  reduce  the  3  cd.  ft.  8  cu.  ft.,  to  the  fraction  of  a  Cd., 
18 


206  COMPOUND  NUMBERS. 

(as  in  373),  and  we  shall  have  j^^g  Cd.  =  -i^g-  Cd.,  which,  reduced  to 
its  equivalent  decimal,  equals  .4375  Cd.,  the  same  as  before.     Hence, 

KuLE.  Divide  the  lowest  denomination  given  hy  that  number 
in  the  scale  which  will  reduce  it  to  the  next  higher  denominationy 
and  annex  the  quotient  as  a  decimal  to  that  higher.  Proceed  in 
the  same  manner  witil  the  whole  is  reduced  to  the  denomination 
required.^    Or, 

Reduce  the  given  numher  to  aj'raction  of  the  required  denomi- 
natioUy  and  reduce  this  fraction  to  a  decimal, 

EXAMPLES    FOR   PRACTICE. 

1.  Eeduce  5  da.  9  h.  46  min.  48  sec.  to  the  decimal  of  a  week. 

Ans.  .7725  wk. 

2.  Reduce  3°  27'  46.44"  to  the  decimal  of  a  sign. 

3.  Reduce  1  R.  11.52  P.  to  the  decimal  of  an  acre. 

4.  What  part  of  4  oz.  is  2  oz.  16  pwt.  19.2  gr  ?      Ans.  .71. 
6.  What  part  of  a  furlong  is  28  rd.  2  yd.  1  ft.  11.04  in.  ? 

6.  Reduce  3|g  to  the  decimal  of  a  pound. 

7.  Reduce  126  A.  4  sq.  ch.  12  P.  to  the  decimal  of  a  town- 
ship. Ans.  .0054893  +  Tp. 

8.  What  part  of  a  fathom  is  3|  ft.  ?  Ans.  .625  fath. 

9.  What  part  of  1\  bushels  is  .45  of  a  peck  ?        Ans.  .09. 

10.  What  part  of  3  A.  2  R.  is  1  R.  11.52  P.?      Ans.  .092. 

11.  Reduce  |  of  ^  of  22|  lb.  to  the  decimal  of  a  short  ton. 

12.  What  part  of  a  f§  is  5  fj  36  T]]^  ?  Ans.  .7  fg. 

13.  Reduce  50  gal.  3  qt.  1  pt.  to  the  decimal  of  a  tun. 

Ans.  .20188  +  T. 
ADDITION. 

S7«5«  Compound  numbers  are  added,  subtracted,  multiplied, 
and  divided  by  the  same  general  methods  as  are  employed  in 
simple  numbers.  The  corresponding  processes  are  based  upon  the 
same  principles ;  and  the  only  modification  of  the  operations  and 
rules  is  that  required  for  borrowing,  carrying,  and  reducing  by  a 
varying,  instead  of  a  uniform  scale. 

376,  1.  What  is  the  sum  of  50  hhd.  32  gal.  3  qt.  1  pt.,  2 
hhd.  19  gal.  1  pt.,  15  hhd.  46|  gal,  and  9  hhd.  39  gal.  2^  qt.  ? 


ADDITION. 


207 


OPERATION.  ^lnalysis.     Writing  the  numbers  so  that 

hhd.    gal.     qt.    pt.         units  of  the  same  denomination  shall  stand 

50     32      3      1  in  the  same  column,  we  add  the  numbers 

2      19      0      1  of  the  right  hand  or  lowest  denomination, 

15     46     1      0  and  find  the  amount  to  be  3  pints,  which  is 

_^| ^^      ^      ^  equal  to  1  qt.  1  pt.  We  write  the  1  pt.  under 

78      11     3      1  the  column  of  pints,  and  add  the  1  qt.  to 

the  column  of  quarts.  The  amount  of  the 
numbers  of  the  next  higher  denomination  is  7  qt.,  which  is  equal  to 
1  gal.  3  qt.  We  write  the  3  qt.  under  the  column  of  quarts,  and  add 
Tlie  1  gal.  to  the  column  of  gallons.  Adding  the  gallons,  we  find  the 
amount  to  be  137  gal.,  equal  to  2  hhd.  11  gal.  Writing  the  11  gal. 
under  the  gallons  in  the  given  numbers,  we  add  the  2  hhd.  to  the 
column  of  hogsheads.  Adding  the  hogsheads,  we  find  the  amount  to 
be  78  hhd.,  which  we  write  under  the  left  hand  denomination,  as  in 
simple  numbers. 

2.  What  is  the  sum  of  ^^^  wk.,  |  da.,  and  |  h. 

oPERATiox.  Analysis.       We 

-5^^  wk.  =r  4  da.  21  h.  36  min.  first  find  the  value 

3    da.   =             14  "  24  min.  ^^  ®^^^  fraction  in 

I   h.  *  ==                        22     "     30  sec.  integers^  of  less  de- 


nominations, (369), 

^         1'^       22  30  and   then   add   the 

Or,  resulting  or  equiva- 

I   da.    X  4  =  /5wk;  lent  compound  num- 

I    h.  X  3j\  X  4  =  -5^  wk ;  bers. 

/^  wk.  +  ^K  wk.  +  ^^^  wk.  =  j  J  wk  ;        ^  Or,    we   may^  re- 


1 J  wk.  =  5  da.  12  h.  22  min.  30  sec.         ^.^^^  *^^  S^'^^^  ^^^^■ 

tions  to  fractions  of 

the  same  denomination,  (368  or  372),  then  add  them,  and  find  the 

value  of  their  sum  in  lower  denominations. 

377.  From  these  examples  and  illustrations  we  derive  the 
following 

Rule.  I.  If  any  of  the  numbers  arc  denominate  fractions^  or 
if  any  of  the  denominations  are  mixed  numhei-Sy  reduce  the  frac- 
tions to  integers  of  lower  denominations. 

II.  Vr^rite  the  numbers  so  that  those  of  the  same  unit  value  will 
stand  in  the  same  column. 

III.  Beginning  at  the  right  hand,  add  each  denomination  as  in 


208  COMPOUND  NUMBERS. 

simple  numbers,  carrying  to  each  succeeding  denomination  one  for 
as  many  units  as  it  takes  of  the  denomination  added,  to  make  one 
of  the  next  higher  denomination. 

Note.  —  The  pupil  cannot  fail  to  see  that  the  principles  involved  in  nddinoj 
compound  numbers  are  the  same  as  those  in  addition  of  simple  numbers;  and 
that  the  only  difference  consists  in  the  different  carrying  unites. 

EXAMPLES    FQR   PRACTICE. 

(1-)  (2-) 

lb.        oz.      pvrt.       gr.  lb.  5        5        9        gf- 

10       8     5  1  8 

7  7  6  2  13 
6     11     7 

21     10  16  . 

12       1     2  2  3 

7  1  19 


14 

6  12 

13 

17 

5   3 

12 

15 

9 

16 

2 

7  15 

20 

13 

2   1 

19 

4 

1   5 

21 

66" 

11   9 

(3.) 

5 

fur. 

rd.   ft. 

in. 

7 

26  11 

9 

4 

16   7 

11 

86  14 

3 

1 

9   2 

8 

5 

10 

1 

6 

2   5 

1 

15  13 

10 

58 

4  5 

(4.) 

2 

19 

JL. 

R.   P. 

sq.  yd. 

sq.ft. 

140 

3  17 

27 

6 

320 

1  SO 

14 

2 

111 

7 

3- 

214 

2  15 

22 

7 

100 

3 

6 

1 

25 

1  36 

8 

104 

2   9 

1 

4 

5.  Add  1  T.  17  cwt.  8  lb.,  5  cwt.  29  lb.  8  oz.,  1  cwt.  42  lb. 
6  oz.,  and  17  lb.  8  oz.  Ans.  2  T.  3  cwt.  97  lb.  6  oz. 

6.  Add  6  yd.  2  ft.,  3  yd.  1  ft.  8  in.,  1  ft.  10  J  in.,  2  yd.  2  ft. 
6i  in.,  2  ft.  7  in.,  and  2  yd.  5  in.  Ans.  16  yd.  2  ft.  1  in. 

7.  Add  4  Cd.  7  cd.  ft.,  2  Cd.  2  cd.  ft.  12  cu.  ft.,  6  cd.  ft.  15 
cu.  ft.,  5  Cd.  3  cd.  ft.  8  cu.  ft.,  and  2  Cd.  1  cu.  ft. 

8.  What  is  the  sum  of  If  hhd.  42  gal.  3  qt.  1}  pt.,  i  gal. 
2  qt.  f  pt.,  and  1.75  pt.  ?  Ans.  2  hhd.  23  gal.  2  qt.  3  gi. 

9.  What  is  the  sum  of  145 J  A.,  7  A.  2  R.  29 J  P.,  1  A.  3  R. 
16.5  P.,  and  |  A.  ?  Ans.  156  A.  39i  P. 

10.  Required  the  sum  of  31  bu.  2  pk.,  10|  bu.,  5  bu.  ^  qt., 
14  bu.  2.75  pk.,  and  |  pk.  Ans.  62  bu.  1  pk.  5  qt.  IJ  pt. 


SUBTRACTIOK.  209 

11.  Required  the  value  of  42  yr.  7-J  mo.  +  10  yr.  3  wk.  5  da. 
+  9|  mo.  +  1  wk.  16  h.  40  min.  +  |  mo.  +  3|  da. 

Ans.  53  yr.  7  mo.  9  da.  23  h.  52  min. 

12.  Add  3  S.  22°  50',  24°  36'  25.7",  17'  18.2",  1  S.  3°  12' 
15.5",  12°  36'  17.8",  and  57.3".  Ans.  6  S.  3°  33'  14.5". 

13.  How  many  units  in  li  gross  7^  doz.,  3  gross  1|  doz.,  |  of 
a  great  gross,  6|  doz.,  and  4  doz.  7  units  ?  Ans.  2183. 

14.  What  is  the  sum  of  240  A.  6  sq.  ch.,  212.1875  sq.  eh., 
and  5  sq.  ch.  10|  P.?  Ans.  262  A.  3  sq.  ch.  13.8  P. 

15.  Add  3|  Pch.  18  cu.  ft.,  84.6  cu.  ft.,  |  Pch.,  and  |«  cu.  ft. 

16.  Add  ?3|,  $25^,  $12|,  S2|,  and  $2.54|.   Ans.  $47.0725. 

17.  What  is  the  sum  of  3  lb  5  g  4  5  2  9  17  gr.,  2  lb  5  5  12 
gr.,  4  §  2  3  1  9  16  gr.  ?  Ans.  5  lb  10  g  4  3  2  9  5  gr. 

18.  AN.  Y.  farmer  received  $.60  a  bushel  for  4  loads  of  corn ; 
the  first  contained  42.4  bu.,  the  second  2866  lb.,  the  third  36i 
bu.,  and  the  fourth  39  bu.  29  lb.  How  much  did  he  receive  for 
the  whole?  Ans.  §100.84-. 

19.  Bought  three  loads  of  hay  at  §8  per  ton.  The  first  weighed 
1.125  T.,  the  second  1|  T.,  and  the  third  2500  pounds;  how  much 
did  the  whole  cost?  Aiis.  $30.20. 

20.  A  man  in  digging  a  cellar  removed  140|  cu.  yd.  of  earth, 
in  digging  a  cistern  24.875  cu.  yd.,  and  in  digging  a  drain  46  cu. 
yd.  20|  cu.  ft.  What  was  the  amount  of  earth  removed,  and  how 
much  the  cost  at  18  cts.  a  cu.  yd.  ? 

Ans.  212.425  cu.  yd.  removed;  $38.24  — cost. 


SUBTRACTION. 
378.   1-  From  18  lb.  5  oz.  4  pwt.  14  gr.  take  10  lb.  6  oz.  10 


pwt.  8  gr. 


I 


Analysis.      Writing   the   subtra- 
hend under   the   minuend,    placing 
units  of  the  same  denomination  under 
each  other,  we  subtract  8  gr.  from 
~Jq        Yi  6  ^^  S^'  ^^^  write  the  remainder,  6 

gr.,  underneath.     Since  we   cannot 
18*  o 


OPERATION. 

•lb. 

oz. 

pwt. 

gr. 

18 

5 

4 

14 

10 

6 

10 

8 

210  COMPOUND  NUMBERS. 

subtract  10  pwt.  from  4  pwt.,  we  add  1  oz.  or  20  pwt.  to  the  4  pwt., 
subtract  10  pwt.  from  the  sum,  and  write  the  remainder,  14  pwt., 
underneath.  Having  added  20  pwt.  or  1  oz.  to  the  6  oz.  in  the  sub- 
trahend, we  find  that  we  cannot  subtract  the  sum,  7  oz.,  from  the  5 
oz.  in  the  minuend ;  we  therefore  add  1  lb.  or  12  oz.  to  the  5  oz.,  sub- 
tract 7  oz.  from  the  sum,  and  write  the  remainder,  10  oz.,  underneath. 
Adding  12  oz.  or  1  lb.  to  the  10  lb.  in  the  subtrahend,  we  subtract 
the  sum,  11  lb.,  from  the  18  lb.  in  the  minuend,  as  in  simple  numbers, 
and  write  the  remainder,  7  lb.,  underneath. 

2.  From  12  bar.  15  gal.  3  qt.  take  7  bar.  18  gal.  1  qt. 

OPERATION.  Analysis.     Proceeding  as  in  the  last 

bar.         gal.         qt  operation,  we  obtain  a  remainder  of  4  bar. 

^^          1^          ^  28J  gal.  2  qt.     But,  J  gal.  =  2  qt.,  which 

added  to  the  2  qt.  in  the  remainder  makes 

^      !:_i__^  1  gal.,  and  this  added  to  the  28  gal.  makes 

4          29  29  gal. ;  and  the  angwer  is  4  bar.  29  gal. 
8.  From  |  of  a  rod  subtract  f  of  a  yard. 

OPERATION, 


Analysis.   "We  first 
find  the  value  of  each 


I  rd.  =  4  yd.  0  ft.  4 '  ....  ,       .      .    . 

3      1    9  a   o      u  fraction  in  integers  ot 

^ lower   denominations, 

'^  1        ^  (369),  and  then  sub- 

Or,  tract    the    less   value 

I  yd.  X~=l  yd.  X  A  =  ^\  rd. ;         ^^^^  ^^'  S^^f  ^^-    ^J' 
0^        ^  *"  ^1        ^^        ^        we    may   reduce    the 

I  rd.  —  2^^  rd.  =  f|  rd. ;  given  fractions  to  frac- 

II  rd.  =  3  yd.  1  ft.  li  in.  tions  of  the  same  de- 

nomination, subtract 
the  less  value  from  the  greater,  and  find  the  value  of  the  remainder 
in  integers  of  lower  denominations. 

379.  From  these  illustrations  we  deduce  the  following 
KuLE.     I.  If  any  of  the  numbers  are  denominate  fractions,  or 
if  any  of  the  denominations  are  mixed  number s,  reduce  the  frac- 
tions to  integers  of  lower  denominations. 

II.  Write  the  subtrahend  under  the  Tninuend,  so  that  units  of  the 
same  denomination  shall  stand  under  each  other. 

III.  Beginning  at  the  right  hand,  subtract  each  denomination 
separately,  as  in  simple  numbers.  . 


I 


SUBTRACTION.  211 


TV.  If  the  numher  of  any  denomination  in  the  siiltrahcnd  ex- 
ceed that  of  the  same  denomination  in  the  minuend j  add  to  the 
mimhcr  in  the  minuend  as  many  units  as  make  one  of  the  next 
higher  denomination^  and  then  subtract ;  in  this  case  add  1  to  the 
next  higher  denomination  of  the  subtrahend  before  subtracting. 
Proceed  in  the  same  manner  with  each  denomination. 


I 


EXAMPLES    FOR    PRACTICE. 

(1.)  (2.) 

mi.         fur.        rd.  ft.         in.  A.  R.  P. 

From  175   3   27   11   4       820  S      26.4 

Take  59   6   10   12   9       150  2   31.86 


Rem.  115   5   16   15   1       170   0   84.54 

(3.)  (4.) 

hhd.      gal.         qt.  yr.       mo.     wk.       da.  .  h. 

5      86      3i                      45      1      8        0      17J 
2      45      Ij  10      9      1      22  6^ 

5.  Subtract  15  rd.  10  ft.  Si  in.  from  26  rd.  11  ft.  8  in. 

Ans.  11  rd.  11  f  in. 

6.  From^l  T.  11  cwt.  30  lbs.  6  oz.  take  18  cwt.  45  lb. 

7.  Subtract  .659  wk.  from  2  wk.  8|  da. 

Ans.  1  wk.  6  da.  5  h.  17  min.  16|  sec. 

8.  From  i^-|  hhd.  take  .90625  gal.  Ans.  32  gal. 

9.  From  f  of  3f  A.  take  8  R.  12.56  P. 

10.  Subtract  ^%  lb.  Troy,  from  10  lb.  8  oz.  8  pwt. 

11.  From  a  pile  of  wood  containing  SQ  Cd.  4  cd.  ft.,  there  was 
sold  10  Cd.  6  cd.  ft.  12  cu.  ft. ;  how  much  remained  ? 

12.  From  5 J  barrels  take  ^  of  a  hogshead. 

Ans.  4  bbl.  11  gal.  1  qt. 

13.  Subtract  |gj  of  a  day  from  f  of  a  week. 

Ans.  4  da.  49  min.  30  sec. 

14.  From. I  of  a  gross  subtract  |  of  a  dozen.        A^is.  6|  dcz. 

15.  From  J  of  a  mile  take  |.^  of  a  rod. 

16.  Subtract  2  A.  8  R.  5.76  P.  from  5  A.  1  R.  24.24  P. 

Ans.  2  A.  2  R.  18.48  P. 


212  COMPOUND  NUMBERS. 

17.  Subtract  .0625  bu.  from  |  pk.  Ans.  4  qt. 

18.  From  the  sum  of  f  of  365|  da.  and  |  of  5|  wk.  take  49^ 
min.  Ans.  33  wk.  1  da.  1  h.  lOf  min. 

19.  From  the  sum  of  §  of  3|  mi.  and  174  ^^-y  ^^^^  ^i  ^^^• 

20.  From  15  bbl.  3.25  gal.  take  14  bbl.  24  gal.  3.54  qt. 

21.  A  farmer  in  Ohio  having  200  bu.  of  barley,  sold  3  loads, 
the  first  weighing  1457  lb.,  the  second  1578  lb.,  and  the  third 
1420  lb. ;  how  many  bushels  had  he  left?     Ans.  107  bu.  9  lb. 

22.  Of  a  farm  containing  200  acres  two  lots  were  reserved,  one 
containing  50  A.  136.4  P.  and  the  other  48  A.  123.3  P.;  the  re- 
mainder was  sold  at  §35  per  acre.     How  much  did  it  bring  ? 

Ans.  $3513.19+. 

23.  An  excavation  58  ft.  long,  37  ft.  wide,  and  6  ft.  deep  is  to 
be  made  for  a  cellar;  after  471  cu. yd.  16  cu.  ft.  972  cu.  in.  of 
earth  have  been  removed,  how  much  more  still  remains  to  be 
taken  out  ?  Ans.     5  cu.  yd.  7  cu.  ft.  756  cu.  in. 

24.  From  the  sum  of  |  lb.,  4|  oz.,  and  31^  pwt.,  take  the  difi*er- 
ence  between  |  oz.  and  |  pwt.     Ans.  1  lb.  3  oz.  8  pwt.  21  gr. 

25.  From  the  sum  of  5j%  A.,  |  of  6|  A.,  -1  E.,  and  ^\  of 

2j\  P.,  take  4  A.  25  P.  12  sq.  yd. 

Ans.  5  A.  3  E.  5  P.  6    sq.  yd. 

88©.   To  find  the  difference  in  dates. 
1.  How  many  years,  months,  days  and  hours  from  3  o'clock  p. 
M.  of  June  15,  1852,  to  10  o'clock  A.  M.  of  Feb.  22,  1860? 

OPERATION.  Analysis.     Since  the  later  of  two  dates 

yr.       mo.    da.      h.  always   expresses  the  greater  period  of 

1860     -J     ^2     10  time,  we  w^rite  the  later  date  for  a  minu- 
IR  5*^     6     1.5     15 

end  and  the  earlier  date  for  a  subtrahend, 

7      8        6      19  placing  the  denominations  in  the  order  of 

the  descending  scale  from  left  to  right, 
(300,  Note  8).  We  then  subtract  by  the  rule  for  subtraction  of 
compound  numbers. 

When  the  exact  numher  of  clays  is  required  for  any  period  not 
exceeding  one  ordinary  year,  it  may  be  readily  found  by  the  fol- 
lowing 


p 


SUBTRACTION. 


TABLE, 


213 


Slioioing  tlie  number  of  days  from  any  day  of  one  month  to  the  same 
day  of  any  other  month  within  one  year. 


FROM  ANT 
DAY  OF 


January 

February 

March 

April 

May 

June 

July 

August 

September 

October 

November 

December 


TO  THE  SAME  DAY  OF  THE  NEXT 


Jan,    Feb.    Mar.   Apr.    May   June  July    Aui,'.  Sept.    Oct.    Nov.    Dec. 


365 

334 

306 

275 

245 

214 

184 

153 

122 

92 

61 

31 


31 
365 
337 
306 
276 
245 
215 
184 
153 
123 
92 
62 


59 
28 
365 
335 
304 
273 
243 
212 
181 
151 
120 
90 


90 
59 
31 
365 
335 
304 
274 
243 
212 
182 
151 
121 


120 

89 

61 

30 

365 

334 

304 

273 

242 

212 

181 

151 


151 

120 

92 

61 

31 

365 

335 

304 

273 

243 

212 

182 


181 

150 

122 

91 

61 

30 

365 

334 

303 

273 

242 

212 


212 

181 

153 

122 

92 

61 

31 

365 

334 

304 

273 

243 


243 

212 

184 

153 

123 

92 

62 

31 

365 

335 

304 

274 


273 

242 

214 

183 

153 

122 

92 

61 

30 

365 

334 

304 


304 

273 

245 

214 

184 

153 

123 

92 

61 

31 

365 

335 


334 

303 

275 

244 

214 

183 

153 

122 

91 

61 

30 

365 


If  the  days  of  the  different  months  are  not  the  same,  the  num- 
ber of  days  of  difference  should  be  added  when  the  earlier  day 
belongs  to  the  month /rom  which  we  reckon,  and  subtracted  when 
it  belongs  to  the  month  to  which  we  find  the  time.  If  the  29th 
of  February  is  to  be  included  in  the  time  computed,  one  day 
must  be  added  to  the  result. 


EXAMPLES   FOR   PRACTICE. 

1.  "War  between  England  and  America  was  commenced  April 
19, 1775,  and  peace  was  restored  January  20,  1783  ]  how  long  did 
the  war  continue  ?  Ans.  7  yr.  9  mo.  1  da. 

2.  The  Pilgrims  landed  at  Plymouth  Dec.  22,  1620,  and  Gen. 
Washington  was  born  Feb.  22,  1732 ;  what  was  the  difference  in 
time  between  these  events  ? 

3.  The  first  settlement  made  in  the  U.  S.  was  at  Jamestown,  Ya., 
May  23,  1607;  how  many  years  from  that  time  to  July  4,  1860  ? 

4.  How  long  has  a  note  to  run,  dated  Jan.  30,  1859,  and  made 
payable  Juno  3,  1861  ?  Ans.  2  yr.  4  mo.  3  da. 


214  COMPOUND   NUMBERS. 

5.  How  many  years,  months,  and  days  from  your  birthday  to 
this  date  ? 

6.  What  length  of  time  elapsed  from  16  minutes  past  10  o'clock, 
A.  M.,  July  4,  1855,  to  22  minutes  before  8  o'clock,  p.  m.,  Dec. 
12,  1860  ?  Alts,  1988  da.  9  h.  22  min. 

7.  What  length  of  time  will  elapse  from  40  minutes  25  seconds 
past  12  o'clock,  noon,  April  21,  1860,  to  4  minutes  36  seconds 
before  5  o'clock,  A.  M.,  Jan.  1,  1862  ? 

8.  How  many  days  from  the  4th  September,  to  the  27th  of 
May  following  ?  Ans.  265  da. 


MULTIPLICATION. 

381.    1.  Multiply  5  mi.  4  fur.  18  rd.  15  ft.  by  6. 

OPERATION.  Analysis.    Writinji:  the  multi- 


5  mi.  4  fur.  18  rd.  15  ft. 
6 


plier  under  the  lowest  denomi- 
nation of  the  multiplicand,  we 
multiply  each   denomination   in 
33  2  33  7i  the   multiplicand    separately   in 

order  from  lowest  to  highest,  as 
in  simple  numbers,  and  carry  from  lower  denominations  to  higher, 
according  to  the  ascending  scale  of  the  multiplicand,  as  in  addition 
of  compound  numbers.     Hence, 

Rule.  I.  Write  the  multiplier  under  the  lowest  denomination 
of  the  midtipUcand. 

II.   Multiply  as  in  simple  numherSj  and  carry  as  in  addition  of 

compound  Jiumhers. 

Notes.  —  1.  "When  the  multiplier  is  large,  and  is  a  composite  number,  we  may 
ehorten  the  work  by  multiplying  by  the  component  factors. 

2.  The  multiplier  must  be  an  abstract  number. 

3.  If  any  of  the  denominations  are  mixed  numbers,  they  may  either  be  re- 
duced to  integers  of  lower  denominations  before  multiplying,  or  they  may  be 
multiplied  as  directed  in  193. 

4.  The  multiplication  of  a  denominate  fraction  is  the  most  readily  performed 
by  193,  after  vyhich  the  product  may  be  reduced  to  integers  of  lower  denomina- 
tions by  369. 

J5§2,  As  the  work  of  multiplying  by  large  prime  numbers  is 
somewhat  tedious,  the  following  method  may  often  be  so  modified 
and  adapted  as  to  greatly  shorten  the  operation. 


MULTIPLICATION. 


215 


1.  How  many  bushels  of  grain  in  47  bags,  eacb  containing  2  bu. 
1  pk.  4  qt.  ? 


I 
I 


FIRST    OPERATION. 

47  =  (5  X  9)  +  2 
2  bu.  1  pk.  4  qt.  X  2 
5 


11  bu.  3  pk.  4  qt.  in    5  bags. 
9 


106  bu.  3  pk.  4  qt.  in  45  bags. 
4    "    3    '<  ^^2     " 

111  bu.  2  pk.  4  qt.   ''  47     " 


SECOND    OPERATION. 

47=(6x  8)  — 1 
2  bu.  1  pk.  4  qt.  X  1 
6 


14  bu.  1  pk.  in    6  bags. 
8 

114  bu.  in  48  bags. 

2    ^^    1  pk.  4  qt.  ''     1  bag. 

Ill  bu.  2  pk.  4  qt.  "  47  bags. 


Analysis.  Multiplying 
the  contents  of  1  bag  by  5, 
and  the  resulting  product  by 
9,  we  have  the  contents  of 
45  bags,  which  is  the  com- 
posite number  next  less  than 
the  given  prime  number,  47. 
Next,  multiplying  the  con- 
tents of  1  bag  by  2,  we  have 
the  contents  of  2  bags,  which, 
added  to  the  contents  of  45 
bags,  gives  us  the  contents 
of  45  +  2  =  47  bags. 

Or,  we  may  multiply  the 
contents  of  1  bag  by  the  fac- 
tors of  the  composite  num- 
ber next  greater  than  the 
given  prime  number,  47,  and 
from  the  last  product  sub- 
tract the  multiplicand. 


EXAMPLES  FOR 

PRACTICE. 

(1-) 

(2-) 

T. 

cwt.    lb. 

OZ. 

mi. 

fur.   rd. 

ft. 

12 

15   27 

9 

14 

6   36 

14 

8 

133 

6   11 

9 

102 

2   20 

8 

10 

(3.) 

(-^0 

A. 

R.    P. 

sq.  yd.  sq.  ft. 

Cd.   cd.  ft. 

cu  ft. 

7 

1   33 

21   7 
6 

10   7 

13 
12 

5.  Multiply  34  bu.  3  pk.  6  qt.  1  pt.  by  14. 

6.  Multiply  4  lb.  10  oz.  18.7  pwt.  by  27. 

Ans.  132  lb.  7  oz.  4.9.  pwt 

7.  Multiply  9  I  3  5  2  9  13  gr.  by  35. 


216  COMPOUND  NUMBERS. 

8.  Multiply  5  gal.  2  qt.  1  pt.  3.25  gi.  by  96. 

9.  Multiply  78  A.  3  R.  15  P.  15  sq.  yd.  by  15f. 

Ans.  1235  A.  1  E.  2  P.  23}  sq.  yd. 

10.  What  is  73  times  9  cu.  yd.  10  cu.  ft.  1424  cu.  in.? 

11.  Multiply  2  lb..  8  oz.  13  pwt.  18  gr.  by  59. 

12.  Multiply  4  yd.  1  ft.  4.7  in.  by  125. 

13.  If  1  qt.  2  gi.  of  wine  fill  1  bottle,  how  much  will  be  re- 
quired to  fill  a  gross  of  bottles  of  the  same  capacity  ? 

14.  Multiply  7  0.  10  f  g  4  f  3  25  Tfl  by  24. 

Ans.  22  Cong.  7  0.  13  f  g  2  f^. 

15.  Multiply  3  hhd.  43  gal.  2.6  gi.  by  17. 

16.  Multiply  9  T.  13  cwt.  1  qr.  10.5  lb.  by  1.7. 

Note. — "NVhen  the  multiplier  contains  a  decimal,  the  multiplicand  may  be  re- 
duced to  the  lowest  denomination  mentioned,  or  the  lower  denominations  to  a 
decimal  of  the  higher,  before  multiplying.  The  result  can  be  reduced  to  the 
compound  number  required.  t^ 

Ans.  16  T.  8  cwt.  2  qr.  20.15  lb. 

17.  If  a  pipe  discharge  2  hhd.  23  gal.  2  qt.  1  pt.  of  w^ter  in  1 
hour,  how  much  will  it  discharge  in  4.8  hours,  if  the  water  flow 
with  the  same  velocity?  Ans.  11  hhd.  25  gal.  1  pt.  2.4  gi. 

18.  What  will  be  the  value  of  1  dozen  gold  cups,  each  cup 
weighing  9  oz.  13  pwt.  8  gr.,  at  $212.38  a  pound  ? 

19.  What  cost  5  casks  of  wine,  each  containing  27  gal.  3  qt.  1 
pt.,  at. $1.37}  a  gallon?  Ans.  $191.64  +  . 

20.  A  farmer  sold  5  loads  of  oats,  averaging  37  bu.  3  pk.  5  qt. 
each,  at  $.65  per  bushel  3  how  much  money  did  he  receive  for  the 
grain?  Ans.  $123.20—. 

DIVISION. 

383.    1.  Divide  37  A.  1  R.  16  P.  by  8. 

OPERATION.  Analysis.    Writing  the  divisor  on 

the  left  of  the  dividend,  we  divide 

^  ;  the  highest  denomination,  and  obtain 

4         2         27  a  quotient  of  4  A.  and  a  remainder 

of  5  A.     Writing  the  quotient  under 

the  deriomination  divided,  we  reduce  the  remainder  to  roods,  making 

20  R.,  vdiich  added  to  the  1  R.  of  the  dividend,  equals  21  R.    Dividing 

this,  we  have  a  quotient  of  2  R.  and  a  remainder  of  5  R.     Writing 


I 


DIVISION. 


217 


the  2  K.  under  the  denomination  divided,  we  reduce  the  remainder  to 
rods,  making  200  P.,  which  added  to  the  16  P.  of  the  dividend,  equals 
216  P.     Dividing  this,  we  have  a  quotient  of  27  P.  and  no  remainder. 
2.  Divide  111  bu.  2  pk.  4  qt.  by  47. 

OPERATION. 

47)lllbu.2pk.4qt.(2bu.lpk.4qt.  a^,,^,,.      The 

94 

divisor  being  large, 

17  bu.  rem.  and  a  prime  num- 

_4  ber,  we  divide  by 

70  pk.  in  17  bu.  2  pk.  long  division,  set- 

47  *i^g      down      the 

Z^     ,  whole  work  of  sub- 

23  pk.  rem.  x      x-  i      i 

o  ^  tracting  and  reduc- 


I 


188  qt.  in  23  pk.  4  qt. 

188 


ing. 


From  these  examples  and  illustrations  we  derive  the  following 
Rule.     I.   Divide  the  highest  denomination  as  in  simple  num- 

hersj  and  each  succeeding  denomination  in  the  same  manner,  if 

there  he  no  remainder, 

II.  If  there  he  a  remainder  after  dividing  any  denomination, 
reduce  it  to  the  next  loicer  denomination,  adding  in  the  given  num- 
her  of  that  denomination,  if  any,  and  divide  as  hefore. 

III.  The  several  partial  quotients  will  he  the  quotient  required. 

KoTES.  —  1.  "When  the  divisor  is  large  and  is  b.  composite  number,  we  may 
shorten  the  work  by  dividing  by  the  factors. 

2.  When  the  divisor  and  dividend  are  both  compound  numbers,  they  must 
both  be  reduced  to  the  same  denomination  before  dividing,  and  then  the  process 
is  the  snme  as  in  simple  numbers. 

3.  The  division  of  a  denominate  fraction  is  most  readily  performed  by  195, 
after  which  the  quotient  may  be  reduced  to  its  equivalent  compound  number, 
by  369.  H  ^  4  F 

EXAMPLES   FOR   PRACTICE. 

(1-)  (2.) 

£  p.         d.        far.  lb.  oz.  pwt.  gr. 

5)62 7 9      3  9)56 6 17 6 

Quotient,   12       9       6       3  61b.    3oz.     8pw.l4gr. 

19 


218  COMPOUND  NUMBERS. 


(3.) 

(4.) 

hhd. 

gal. 

qt. 

pt. 

T. 

cwt. 

qr. 

lb. 

12)9 

28 

2 

19 

)873 

19 

2 

4 

49      2       1'  19       13       2       16 

6.  Divide  858  A.  1  R.  17  P.  6  sq.  yd.  2  sq.  ft.  by  7. 

Ans.  51  A.  31  P.  8  sq.  ft. 

6.  Divide  192  bu.  3  pk.  1  qt.  1  pt.  by  9. 

7.  Divide  336  yd.  4  ft.  3^  in.  by  21.       Aiis.  16  yd.  2^  in. 

8.  Divide  77  sq.  yd.  5  sq.  ft.  82  sq.  in.  by  13. 

Ans.  5  sq.  yd.  8  sq.  ft.  106  sq.  in. 

9.  Divide  678  cu.  yd.  1  cu.  ft.  1038.05  cu.  in.  by  67. 

10.  Divide  1986  mi.  3  fur.  20  rd.  1  yd.  by  108. 

11.  Divide  12  sq.  mi.  1  R.  30  P.  by  22^. 

Ans.  341  A.  1  E.  16|  P. 

Note  4. — Observe  that  22^  =  ^  ;  hence,  multiply  by  2,  and  divide  the  result 
toy  45. 

12.  Divide  365  da.  6  h.  by  240. 

13.  Divide  3794  cu.  yd.  20  cu.  ft.  7091  cu.  in.  by  331. 

14.  Divide  121  lb.  3g  2^  19  4  gr.  by  13|. 

15.  Divide  28°  51'  27.76*5''  by  2.754.      Ans.  10°  28'  42^. 

16.  Divide  202  yd.  1  ft.  6|  in.  by  f . 

Ans.  337  yd.  1  ft.  7i  in. 

17.  Divide  1950  da.  15  h.  15|  min.  by  100. 

18.  If  a  town  4  miles  square  be  divided  equally  into  124  farms, 
how  much  will  each  farm  contain  ?         Ans.  82  A.  2  R.  12||  P. 

19.  A  cellar  48  ft.  6  in.  long,  24  ft.  wide,  and  65  ft.  deep,  was 
excavated  by  6  men  in  8  days;  how  many  cubic  yards  did  each 
man  excavate  daily?  Ans.  5  cu.  yd.  22  cu.  ft.  1080  cu.  in. 

20.  How  many  casks,  each  containing  2  bu.  3  pk.  6  qt.,  can  be 
filled  from  356  bu.  3  pk.  5  qt.  of  cherries?  Ans.  121  J. 


LONGITUDE   AND   TIME. 

384.  Since  the  earth  performs  one  complete  revolution  on  its 
axis  in  a  day  or  24  hours,  the  sun  apj^ears  to  j^ciss  from  east  to 
west  round  the  earth,  or  through  360°  of  longitude  once  in  every 


LONGITUDE  AND  TIME.  219 

24  hours  of  time.     Hence  the  relation  of  time  to  the  real  motion 
of  the  earth  or  the  a2'>parcnt  motion  of  the  sun^  is  as  follows: 

Time.  ,        Longitude. 

24  h.  =     360° 

1  h.  or  60  min.  =    ^^T      =     15°    =    900^ 

1  min.  or  60  sec.    ==   ^^°  =    Vn  ^       =     15^    =    QOO''^ 

Isec.                       =    ir  =    'i'r'     -     15^" 

Hence,  1  h.  of  time    =    15°  of  longitude. 

1  min.     "  =    15^  '' 

^  1  sec.      '*  =    15^/  **           " 

CASE  I. 

385.  To  find  the  difference  of  longitude  between  twc 
places  or  meridians,  when  the  difi:erence  of  time  is 
known. 

Analysis.  A  difference  of  1  h.  of  time  corresponds' to  a  difference 
of  15°  of  longitude,  of  1  min.  of  time  to  a  difference  of  15^  min.  of 
longitude,  and  of  1  sec.  of  time  to  a  difference  of  15^^  of  longitude, 
(384).     Hence,  the 

Rule.  Multiply  the  difference  of  time,  expressed  in  lirntrs^ 
minutes^  and  seconds,  hy  15,  according  to  the  rule  for  midtipUca- 
tion  of  compound  numbers  ;  the  product  will  he  the  difference  of 
longitude  in  degrees,  minuteSj  and  seconds. 

Notes.  —  1.  If  one  place  be  in  enst,  and  the  other  in  west  lonoritude,  the  dif- 
ference of  longitude  is  found  by  udding  them,  and  if  the  sum  be  greater  than 
180°,  it  must  be  subtracted  from  360°. 

2.  Since  the  sun  appears  to  move  from  east  to  west,  when  it  is  exactly  12 
o'clock  at  one  place,  it  will  he  pn^t  12  o'clock  at  all  places  east,  and  hefore  12  at 
all  places  west.  Hence,  if  the  difference  of  time  between  two  places,  be  HuhirHvird 
from  the  time  at  the  easterly  place,  the  result  will  be  the  time  at  the  westerly 
place ;  and  if  the  difference  be  added  to  the  time  at  the  westerly  place  the  result 
will  be  the  time  at  the  easterly  place. 

EXAMPLES    FOR   PRACTICE. 

1.  When  it  is  9  o'clock  at  Washington,  it  is  8  h.  7  min.  4  sec. 
at  St.  Louis;  the  longitude  of  Washington  heing  77°  1',  west, 
what  is  the  longitude  of  St.  Louis  ?  Ans.  90°  15'  west. 

2.  The  sun  rises  at  Boston  1  h.  11  min.  56  sec.  sooner  than 
at  New  Orleans;  the  longitude  of  New  Orleans  being  80°  2'  west, 
what  is  the  longitude  of  Boston?  Ans.  71°  3'  west. 


I 


220  COMPOUND  NUMBERS. 

3.  When  it  is  half  past  2  o'clock  in  the  morning  at  Havana,  it 
is  9  h.  13  min.  20  sec.  a.  m.  at  the  Cape  of  Good  Hope;  the 
longitude  of  the  latter  place  being  18°  28'  east,  what  is  the 
longitude  of  Havana  ?  Ans.  82°  22'  west. 

4.  The  difference  of  time  between  Valparaiso  and  Rome  is  6  h. 
8  min.  28  sec. ;  what  is  the  difference  in  longitude  ? 

6.  A  gentleman  traveling  East  from  Fort  Leavenworth,  which 
is  in  94°  44'  west  longitude,  found,  on  arriving  at  Philadelphia,  that 
his  watch,  an  accurate  time  keeper,  was  1  h.  18  min.  16  sec.  slower 
than  the  time  at  Philadelphia ;  what  is  the  longitude  of  Philadel- 
phia ?  Ans,  75°  10'  west. 

6.  When  it  is  12  o'clock  M.  at  San  Francisco  it  is  3  h.  58  min. 
23  J  sec.  p.  M.  at  Rochester,  N.  Y;  the  longitude  of  the  latter 
place  being  77°  51'  W.,  what  is  the  longitude  of  San  Francisco? 

7.  A  gentleman  traveling  West  from  Quebec,  which  is  in,  71° 
12'  15"  W.  longitude,  finds,  on  his  arrival  at  St.  Joseph,  that  his 
watch  is  2  h.  33  min.  53j|  sec.  faster  than  true  time  at  the  latter 
place.  If  his  watch  has  kept  accurate  time,  what  is  the  longitude 
of  St.  Joseph  ?  Ans.  109°  40'  44"  V. 

8.  A  ship's  chronometer,  set  at  Greenwich,  points  to  5  h.  40 

min.  20  sec.  p.  M.,  when  the  sun  is  on  the  meridian ;  what  is  the 

ship's  longitude  ?  Ans.  85°  5'  west. 

Note  .3. — Greenwich,  Eng.,  is  on  the  meridian  of  0°,  and  from  this  meridian 
longitude  is  reckoned. 

9.  The  longitude  of  Stockholm  being  18°  3'  30"  E.,  when  it  is 
midnight  there,  it  is  5  h.  51  min.  41 1  sec.  A.  M.  at  New  York;  what 
is  the  longitude  of  New  York  from  Greenwich  ? 

Ans,  74°  1'  6"  W. 

10.  A  vessel  set  sail  from  New  York,  and  proceeded  in  a  south- 
easterly direction  for  24  days.  The  captain  then  took  an  obser- 
vation on  the  sun,  and  found  the  local  time  at  the  ship's  meridian 
to  be  10  h.  4  min.  36.8  sec.  A.  M. ;  at  the  moment  of  the  observa- 
tion, his  chronometer,  which  had  been  set  for  New  York  time, 
showed  8  h.  53  min.  47  sec.  a.  m.  Allowing  that  the  chronometer 
had  gained  3.56  sec.  per  day,  how  much  had  the  ship  changed  her 
lon«;itude  since  she  set  sail?  Ans.    18°  3'  48.6". 


LONGITUDE  AND  TIME.  221 


CASE   II. 

38G.  To  find  the  difference  of  time  between  two 
places  or  meridians,  when  the  difference  of  longitude 
is  known. 

Analysis.  A  difference  of  15°  of  longitude  produces  a  difference 
of  1  h.  of  time,  15^  of  longitude  a  difference  of  1  min.  of  time,  and 
15^^  of  longitude  a  difference  of  1  sec.  of  time,  (384).     Hence  the 

Rule.  Divide  the  difference  of  longitude^  expressed  in  degrees, 
minutes,  and  seconds,  hi/  15,  according  to  the  rule  for  division:  of 
compound  numbers;  the  quotient  will  he  the  difference  of  time  in 
hours,  minutes,  and  seconds, 

EXAMPLES    FOR    PRACTICE. 

1.  Washington  is  77°  1'  and  Cincinnati  is  84°*24'  west  longi- 
tude; what  is  the  difference  of  time?        Ans.  29  min.  32  see. 

2.  Paris  is  2°  20'  and  Canton  113°  14'  east  longitude;  what  is 
the  difference  in  time  ? 

3.  Buffalo  is  78°  55'  west,  and  the  city  of  Rome  20°  30'  east 
longitude ;  what  is  the  difference  in  time  ? 

Ans.  6  h.  37  min.  40  sec. 
^      4.  A  steamer  arrives  at  Halifax,  C3°  36'  west,  at  4  h.  30  min. 
p.  M.;  the  fact  is  telegraphed  to  New  York,  74°  1'  west,  without 
loss  of  time ;  what  is  the  time  of  its  receipt  at  New  York  ? 

5.  The  longitude  of  Cambridge,  Mass.,  is  71°  7'  west,  and  of 
Cambridge,  England,  is  5'  2"  east;  what  time  is  it  at  the  former 
place  when  it  is  12  M.  at  the  latter  ? 

i      .  Ans.   7  h.  15  min.  11||  sec.  A.  M. 

6.  The  longitude  of  Pekin  is  118°  east,  and  of  Sacramento 
City  120°  west;  what  is  the  difference  in  time? 

f       7.  The  longitude  of  Jerusalem  is  35°  32'  east,  and  that  of 
Baltimore  76°  37'  west;  when  it  is  40  minutes  past  6  o'clock 

,  A.  M.  at  Baltimore,  what  is  the  time  at  Jerusalem? 

I      8.  What  time  is  it  in  Baltimore  when  it  is  6  o'cfeck  p.  M.  at 
Jerusalem?  Ans.  10  h.  31  min.  2#;Bec.  A.M. 

19  * 


222  COMPOUXD  NUMBEKS. 

9.  The  longitude  of  Springfield,  Mass.,  is  72°  85'  45"  W.,  and 
of  Galveston,  Texas,  94°  46'  34"  W.;  when  it  is  20  min.  past  6 
o'clock  A.  M.  at  Springfield,  what  time  is  it  at  Galveston  ? 

10.  The  longitude  of  Constantinople  is  28°  49'  east,  and  of  St. 
Paul  93°  5'  Vv^est;  when  it  is  3  o'clock  p.  M.  at  the  latter  place, 
what  time  is  it  at  the  former  ? 

11.  What  time  is  it  at  St.  Paul  when  it  is  midnight  at  Constan- 
tinople? Ans.  3  h.  52  min.  24  sec.  p.  M. 

12.  The  longitude  of  Cambridge,  Eng.,  is  5'  2"  E.,  and  of 
Mobile,  Ala.,  88°  1'  29"  W.;  when  it  is  12  o'clock  M.  at  Mobile, 
what  is  the  time  at  Cambridge  ? 

PEOMISCUOUS    EXAMPLES    IN    COMPOUND    NUMBERS. 

1.  In  9  lb.  S^  1^  29  19  gr.  how  many  grains? 

2.  IIow  much  will  3  cwt.  12  lb.  of  hay  cost,  at  §15^  a  ton? 

3.  In  27  yd.  2  qr.  how  many  Eng.  ells?  Aas.   22. 

4.  Heduce  §18.945  to  sterling  money.    Ans.  £3  18s.  3yyjd. 

5.  In  4  yr.  48  da.  10  h,  45  sec.  how  many  seconds  ? 

6.  How  many  printed  pages,  2  pages  to  each  leaf,  will  there  be 
in  an  octavo  book  having  24  fully  printed  sheets?      Aus.  384. 

7.  At  1/6  sterling  per  yard,  how  many  yards  of  cloth  may  be 
bought  for  £5  Gs.  Gd.  ?  Ans.  71  yd. 

8.  In  4  mi.  51  ch.  73  1.  how  many  links? 

9.  In  22  A.  3  II.  33  sq.  rd.  2|  sq.  yd.  how  many  square  yards  ? 

10.  IIow  many  demijohns,  each  containing  3  gal.  1  qt.  1  pt., 
can  be  filled  from  3  hhd.  of  currant  wine  ?  Ans.  56. 

11.  Paid  §375.75  for  2^  tons  of  cheese,  and  retailed  it  at  9| 
cts.  a  pound ;  how  much  was  my  whole  gain  ? 

12.  A  gentleman  sent  a  silver  tray  and  pitcher,  weighing  3  lb. 
9  oz.,  to  a  jeweler,  and  ordered  them  made  into  tea  spoons,  each 
weighing  1  oz.  5  pwt. ;  how  many  spoons  ought  he  to  receive? 

Ans.  3  doz. 

13.  What  part  of  4  gal.  3  qt.  is  2  qt.  1  pt.  2  gi.  ?       Ans.  4|. 

14.  lleduce  |  of  j\  of  a  rod  to  the  fraction  of  yard. 

15.  How  many  yards  of  carpeting  1  yd.  wide,  will  be  required 
to  cover  a  floor  26  J  ft.  long,  and  20  ft.  wide  ?  Ans.  58|. 


/ 


PEOMISCUOUS   EXAMPLES.  223 

16.  If  I  purchase  15  T.  3  cwt.  3  qr.  24  lb.  of  English  iron,  by 
long  ton  weight,  at  6  cents  a  pound,  and  sell  the  same  at  ^140  per 
short  ton,  how  much  will  I  gain  by  the  transaction  ? 

17.  What  will  be  the  expense  of  plastering  a  room  40  ft.  long, 
36 J  ft.  wide,  and  22i  ft.  high,  at  18  cents  a  sq.  yd.,  allowing  1375 
sq.  ft.  for  doors,  windows,  and  base  board?  Ans.  $69.78 J. 

18.  How  much  tea  in  23  chests,  each  weighing  78  lb.  0  oz.  ? 

19.  Valparaiso  is  in  latitude  33°  2'  south,  and  Mobile  30°  41' 
north ;  what  is  their  difference  of  latitude  ?  Ans.  63°  43'. 

20.  If  a  druggist  sell  1  gross  4  dozen  bottles  of  Congress  water 
a  day,  how  many  will  he  sell  during  the  month  of  July  ? 

21.  Eighteen  buildings  arc  erected  on  an  acre  of  ground,  each 
occupying,  on  an  average,  4  sq.  rd.  120  sq.  ft.  84  sq.  in. ;  how 
much  ground  remains  unoccupied  ? 

22.  At  $13  per  ton,  how  much  hay  may  be  bought  for  $12.02  J  ? 

23.  If  1  pk.  4  qt.  of  wheat  cost  $.72,  how  much  will  a  bushel 
cost?  Ans.  $1.92. 

24.  IIow  many  bushels,  Indiana  standard,  in  36244  lbs.  of 
wheat  ? 

25.  At  20  cents  a  cubic  yard,  how  much  will  it  cost  to  dig  a 
cellar  32  ft.  long,  24  ft.  wide,  and  6  ft.  deep?    Ans.  $34.13  +  . 

26.  If  the  wall  of  the  same  cellar  be  laid  IJ  feet  thick,  what 
will  it  cost  at  $1.25  a  perch  ?  Ans,  $50.90  jf. 

27.  The  forward  wheels  of  a  wagon  are  10  ft.  4  in.  in  circum- 
ference, and  the  hind  wheels  15 J  ft.;  how  many  more  times  will 
the  forward  wheels  revolve  than  the  hind  wheels  in  running  from 
Boston  to  N.  Y.,  the  distance  being  248  miles?        Aois.  42240. 

28.  Bought  15  cwt.  22  lb.  of  rice  at  $3.75  a  cwt,  and  7  cwt. 
36  lb.  of  pearl  barley  at  $4.25  a  cwt.  How  much  would  be  gained 
by  selling  the  whole  at  4}  cents  a  pound  ?  Ans.  $13,255. 

29.  From  f  of  3  T.  10  cwt.  subtract  -^^  of  7  T.  3  cwt.  26  lb. 

30.  What  is  the  value  in  avoirdupois  weight  of  16  lb.  5  oz.  10 
pwt.  13  gr.  Troy?  Ans.  13  lb.  8  oz.  11.4+dr. 

31.  What  decimal  of  a  rod  is  1  ft.  7.8  in.  ? 

32.  If  a  piece  of  timber  be  9  in.  wide  and  6  in.  thick,  what 
length  of  it  will  be  required  to  make  3  cu.  ft.  ?  Ans  8  ft. 


224  COMPOUND  NUMBERS. 

33.  If  a  board  be  16  in.  broad,  what  length  of  it  will  make  7 
sq.  ft.  ?  Ans.  5i  ft. 

34.  If  a  hogshead  contain  10  cubic  feet,  how  many  more  gal- 
lons of  dry  measure  will  it  contain  than  of  beer  measure  ? 

35.  How  many  tons  in  a  stick  of  hewn  timber  60  ft.  long,  and  1 
ft.  9  in.  by  1  ft.  1  in.  ?  Ans.  2.275  tons. 

36.  Subtract  -|  bu.  +  |  of  f  f  of  3i  qt.  from  5  bu.  3|^  qt. 

ot 

Ans.  161  pk. 

37.  What  is  the  difference  between  f  of  5  sq.  mi.  250  A.  3  R., 
and  3i  times  456  A.  3  R.  14  P.  25  sq.  yd.  ? 

Ans.  2  sq.  mi.  254  A.  2  R.  26  P.  24|  sq.  yd. 

38.  How  many  pounds  of  silver,  Troy  weight,  are  equivalent 
in  value  to  5.6  lb.  of  gold  by  the  English  government  standard  ? 

Ans.  80  lb.  2  pwt.  19.2768  gr. 

39.  If  a  piece  of  gold  is  |  pure,  how  many  carats  fine  is  it  ? 

40.  In  gold  16  carats  fine  what  part  is  pure,  and  what  part  is 
alloy  ? 

41.  A  man  having  a  piece  of  land  containing  384|  A.,  divided 
it  between  his  two  sons,  giving  to  the  elder  22  A.  1  R.  20  P.  more 
than  to  the  younger ;  how  many  acres  did  he  give  to  each  ? 

Ans.  203  A.  2  R.  14  P.,  elder ;  181  A.  0  R.  34  P.,  younger. 

42.  4000  bushels  of  corn  in  Illinois  is  equal  to  how  many  bushels 
in  New  York  ?  Ans.  3586^^  bu. 

43.  The  market  value  being  the  same  in  both  States,  a  farmer 
in  New  Jersey  exchanged  110  bu.  of  cloverseed,  worth  $4  a 
bushel,  with  a  farmer  in  New  York  for  corn,  worth  M  a  bushel, 
which  he  sold  in  his  own  State  for  cash.  The  exchange  being 
made  by  weight,  in  whose  favor  was  the  difference,  and  how  much 
in  cash  value  ? 

Ans.  The  N.  J.  farmer  gained  694  ^^-  corn,  worth  S46/y. 

44.  The  great  pyramid  of  Cheops  measures  763.4  feet  on  each 
side  of  its  base,  which  is  square.    How  many  acres  does  it  cover  ? 

45.  The  roof  of  a  house  is  42  ft.  long,  and  each  side  20  ft.  6 
in.  wide;  what  will  the  roofing  cost  at  $4. 62 J  a  square  ? 


/ 


I 


PROMISCUOUS   EXAMPLES.  225 

46.  If  17  T.  15  cwt.  62}  lb.  of  iron  cost  $1833.593,  how  much 
will  1  ton  cost? 

47.  How  many  wine  gallons  will  a  tank  hold,  that  is  4  ft.  long 
by  iij  it.  wido,  and  If  i't.  deep?  A)ts.  1871^^^  gal. 

48.  What  will  be  the  cost  of  300  bushels  of  wheat  at  9s.  4d. 
per  bushel,  31ichigan  currency  ?  Ans.  $350. 

49.  What  will  be  the  cost  in  Missouri  currency? 

50.  W^hat  will  be  the  cost  in  Delaware  currency  ?  ' 

51.  What  will  be  the  cost  in  Georgia  currency?    Ans.  $600. 

52.  What  will  be  the  cost  in  Canada  currency?     A.ns,  $560. 
^        53.  Bought  the  following  bill  of  goods  in  Boston  : 

/""  6 J  yd.  Iriph  linen      @  5/4 

12       "    flannel  ''   3/9 

8}    "    calico  ''    1/7 

9      ''    ribbon  ''     /9 , 

4}  lb.  coffee  ^-    1/5 

6 1  gal.  molasses  "    3/8 

W^hat  was  the  amount  of  the  bill  ?  Ans.   $21.76  +. 

54.  How  many  pipes  of  Madeira  are  equal  to  22  pipes  of 
sherry  ? 

55.  A  cubic  foot  of  distilled  water  weighs  1000  ounces  avoirdu- 
pois; what  is  the  weight  of  a  wine  gallon  ?      Ans.   8  lb.  5-^|  oz. 

56.  There  is  a  house  45  feet  long,  and  each  of  the  two  sides 
of  the  roof  is  22  feet  wide.  Allowing  each  shingle  to  be  4  inches 
wide  and  15  inches  long,  and  to  lie  one  third  to  the  weather,  how 
many  half-thousand  bunches  will  be  required  to  cover  the  roof? 

Ans.  28^^/3. 

57.  A  cistern  measures  4  ft.  6  in.  square,  and  6  ft.  deep;  how 
many  hogsheads  of  water  will  it  hold  ? 

58.  If  the  driving  wheels  of  a  locomotive  be  18  ft.  9  in.  in  cir- 
cumference, and  make  3  revolutions  in  a  second,  how  long  will  the 
locomotive  be  in  running  150  miles  ? 

Ans.  3  h.  54  min.  40  sec. 
9^   59    In  traveling,  when  I  arrived  at  Louisville  my  watch,  which 
was  exactly  right  at  the  beginning  of  my  journey,  and  a  correct 

p 


226  COMPOUND  NUMBERS. 

timekeeper,  was  1  h.  6  min.  52  sec.  fast;  from  what  direction 
had  I  come,  and  how  far?  A)ts.  From  the  east,  16°  43'. 

60.  How  many  U.  S.  bushels  will  a  bin  contain  that  is  8.5  ft. 
long,  4.25  ft.  wide,  and  3 J  ft.  deep? 

61.  Eeduce  3  hhd.  9  gal.  3  qt.  wine  measure  to  Imperial  gal- 
lons. Ans,  165.5807+  Imp'l  gal. 

62.  A  man  owns  a  piece  of  land  which  is  105  ch.  85  1.  long, 
and  40  ch.  15  1.  wide;  how  many  acres  does  it  contain  ? 

63.  A  and  B  own  a  farm  together;  A  owns  -^^  of  it  and  B  the 
remainder,  and  the  difference  between  their  shares  is  15  A.  1  R. 
28}  P.     How  much  is  B^s  share?  Ans.  38  A.  2  R.  11|  P. 

64.  At  83.40  per  square,  what  will  be  the  cost  of  tinning  both 
sides  of  a  roof  40  ft.  in  length,  and  whose  rafters  are  20  ft.  6  in. 
long?  Ans.  $55.76. 

65.  What  is  the  value  of  a  farm  189.5  rd.  long  and  150  rd. 
wide,  at  S3  If  per  acre?         /     /,,    . 

GQ.  Eeduce  9.75  tons  of  liewn  timber  to  feet,  board  measure, 
that  is,  1  inch  thick.  Ans.   5850  ft. 

67.  How  many  wine  gallons  will  a  tank  contain  that  is  4  ft. 
long,  3-J  ft.  wide,  and  2f  ft.  deep?  ^/?.^.   299i|  gal. 

68.  If  a  load  of  wood  be  12  ft.  long,  and  3  ft.  6  in.  wide,  how 
high  must  it  be  to  make  a  cord  ?    ^^  . ,  ,  -"; 'rv      ^  *   L.  >7^  / 

69.  In  a  school  room  32  ft.  long,  18  ft.  wide,  and  12  ft.  6  in. 
high,  are  60  pupils,  each  breathing  10  cu.  ft.  of  air  in  a  minute. 
In  how  long  a  time  will  they  breathe  as  much  air  as  the  room 
contains  ?  )  / 

70.  A  man  has  a  piece  of  land  201|  rods  long  and  4H  rods 
w^ide,  which  he  wishes  to  lay  out  into  square  lots  of  the  greatest 
possible  size.     How  many  lots  will  there  be  ?  Ans.  396. 

71.  A  man  has  4  pieces  of  land  containing  4  A.  3  B.  20  P.,  6 
A,  3  II.  12  P.,  9  A.  3  B.,  and  11  A.  2  B.  32  P.  respectively. 
It  is  required  to  divide  each  piece  into  the  largest  sized  building 
lots  possible,  each  lot  containing  the  same  area,  and  an  exact  num- 
ber of  square  rods.     How  much  land  will  each  lot  contain  ? 

A71S,   156  P 


DUODECIMALS. 


22T 


DUODECIMALS. 

•18 T.  Duodecimals  are  the  parts  of  a  unit  resulting  from  con- 
tinually dividing  by  12;  as  1,  -^^^  -j\^,  T72H'  ^^^-  ^^  practice, 
duodecimals  are  applied  to  the  measurement  of  extension,  the  foot 
being  taken  as  the  unit. 

In  the  duodecimal  divisions  of  a  foot,  the  different  orders  of 
units  are  related  as  follows : 

1^     (inch  or  prime) is   J^    of  afoot,  or  1  in.  linear  measure. 

y^  (second)or  y'2  of  ^2? "   y^^^-  of  a  foot,  or  1  ''    square       '* 

r^^(third)ory'2of  j'aof  3^2-,..  *' jJ^^ofafoot,  orl  ''    cubic 

TABLE. 

12  fourths,  [''''),  make  1  third V^ 

12  thirds  ''      1  second,  .......    V 

12  seconds  "      Iprime, ... V 

12  primes,  "      1  foot, ft. 

Scale  —  uniformly  12. 

The  marks  ^,  ^^,  ^^^,  ^^^^,  are  called  indices. 

Notes.  —  1.  Duodecimals  are  really  common  fractions,  and  can  always  be 
treated  as  such  ;  but  usually  their  denominators  are  not  expressed,  and  they  are 
treated  as  compound  numbers. 

2.  The  word  duodecimnl  is  derived  from  the  Latin  term  duodecim,  signifying 
12. 

ADDITION  AND  SUBTRACTION. 
388.    Duodecimals  are   added  and  subtracted  in  the  same 
manner  as  compound  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  Add  12  ft.  r  8",  15  ft.  3'  5",  17  ft.  9'  7". 

Ans.   45  ft.  8'  8''. 

2.  Add  136  ft.  11'  6"  8''',  145  ft.  10'  8"  5'",  160  ft.  9'  5"  5'". 

Ans,   443  ft.  7'  8"  6'". 

3.  From  36  ft.  7'  11"  take  12  ft.  9'  5".  Ans.  23  ft.  10'  6". 

4.  A  certain  room  required  300  sq.  yd.  2  sq.  ft.  5'  of  plastering. 
The  walls  required  50  sq.  yd.  1  sq.  ft.  7'  4",  62  sq.  yd.  5'  3",  48 
sq.  yd.  2  sq.  ft.,  and  42  sq.  yd.  2  sq.  ft.  3'  4",  respectively.  Re- 
quired the  area  of  the  ceiling.       Ans.  97  sq.  yd.  5  sq.  ft.  1'  1". 


OPERATION. 
9  ft.        8' 

4  ft.     7' 

5  ft. 
38  ft. 

7' 
8' 

8" 

228  DUODECIMALS. 

MULTIPLICATION. 

3§0«  In  the  multiplication  of  duodecimals,  the  product  of  two 
dimensions  is  area,  and  the  product  of  three  dimensions  is 
solidity  (282,  286). 

We  observe  that 

V  X  Ift.^:^^^  of  1ft.  =K 

V  X  1  ft.  ^  jl:f  of  1  ft.  =  V^. 
y   XV     =  -iV  X  ,L  of  1  ft.   =  v. 

1//  X  1^     =  jh  X  A  of  1  ft.  =  y.     Hence, 
The  product  of  any  two  orders  is  of  the  order  denoted  by  the 
sum  of  their  indices. 

390.    1.  Multiply  9  ft.  8'  by  4  ft.  T, 

Analysis.  Beginning  at  the  right, 
8^  X  7^  =  56^^  =  4^  8^^ ;  writing 
the  8^^  one  place  to  the  right,  we  re- 
serve the  4^  to  be  added  to  the  next 
product.  Then,  9  ft.  X  7^  +  4^  = 
07^  =  5  ft.  7^,  which  we  write  in  the 
44  ft.      3'      8",  Ayis,  places  of  feet  and  primes.     Next  mul- 

tiplying by  4  ft.,  we  have  8^  X  4  ft. 
=  32^  =  2  ft.  8^ ;  writing  the  8^  in  the  place  of  primes,  we  reserve  the 
2  ft.  to  be  added  to  the  next  product.  Then,  9  ft.  X  4  ft.  +  2  ft.  = 
38  ft.,  w^hich  we  write  in  the  place  of  feet.  Adding  the  partial  pro- 
ducts, we  have  44  ft.  y  8^^  for  the  product  required.     Hence  the 

Rule.  I.  Write  the  several  terms  of  the  muliijplier  under  the 
corresponding  terms  of  the  multiplicand, 

II.  Multiply  each  term  of  the  multiplicand  hy  each  term  of  the 
multiplier y  heginning  icith  the  loicest  term  in  each,  and  call  the  pro- 
duct  of  any  two  orders,  the  order  denoted  hy  the  sum  of  their  in- 
dices, carrying  1  for  every  12. 

III.  Add  the  partial  products ;  their  sum  will  he  the  required 
answer. 

EXAMPLES    FOR   PRACTICE. 

1.  How  many  square  feet  in  a  floor  16  ft.  8'  wide,  and  18  ft.  5' 
long? 

2.  How  much  wood  in  a  pile  4  ft.  wide,  3  ft.  8'  high,  and  23  ft 
7Mon<?? 


I 


MULTIPLICATION.  229 

|8.  ir  a  floor  be  79  11.  8'  by  88  ft.  11',  how  many  square  yards 
Joes  it  contain  ?  Aus.  344  yd.  4  ft.  4'  4". 

4.  If  a  block  of  marble  be  7  ft.  6'  long,  3  ft.  3'  wide,  and  1  ft. 
10'  thick,  what  are  the  solid  contents?  Ans.  44  ft.  8'  3", 

5.  How  many  solid  feet  in  7  sticks  of  timber,  each  56  ft.  long, 
11  inches  wide,  and  10  inches  thick?  Ans.  299  ft.  5'  4". 

6.  How  many  feet  of  boards  will  it  require  to  inclose  a  building 
60  ft.  6'  long,  40  ft.  8'  wide,  22  ft.  high,  and  each  side  of  the 
roof  24  ft.  2',  allowing  523  ft.  3'  for  the  gables,  and  making  no 
deduction  for  doors  and  windows  ?  Ans,  7880  ft.  5'. 


CONTRACTED    METHOD. 


391.  The  method  of  contracting  the  multiplication  of  deci- 
mals may  be  applied  to  duodecimals,  the  only  modification  being 
in  carrying  according  to  the  duodecimal,  instead  of  the  decimal, 
scale. 

1.  Multiply  7  ft.  3'  5"  8'"  by  2  ft.  4'  7"  9'",  rejecting  all  de- 

i nominations  below  seconds  in  the  product. 
OPERATION.  Analysis.     ^Ye  write   2  ft.,  the 

7  ft.  3'  b"  8'"  units  of  the  multiplier,  under  the 

9''^  7''     4'  2  ft  lowest  order  to  be  reserved  in  the 

14  ft.  6'   IV  product,  and  the  other  terms  at  the 

K  2  ft.   5'     2"  l^ft,  with  their  order  reversed.  Then 

B  4'     3"  it  is   obvious  that  the   product  of 

K  5"  each  term  by  the  one   above  it  is 

B  17  ft    4'     9"dtz   Ans.      seconds.     Hence  we  multiply  each 

term  of  the  multiplier  into  the  terms 
above  and  to  the  left  of  it  in  the  multiplicand,  carrying  from  the 
rejected  terms,  thus ;  in  multiplying  by  2  ft.,  we  have  8^^^  X  2  ft.  = 
IG^^^  =  1^^  4^^^,  which  being  nearer  1^^  than  2^^,  gives  1^^  to  be  car- 
ried to  the  first  contracted  product.  In  multiplying  by  4^,  we  have 
15//  X  4^  =  20^^^^  =  V  S'^^,  which  being  nearer  2''  than  V,  gives 
2^^  to  be  carried  to  the  second  contracted  product,  and  so  on. 


EXAMPLES    FOR   PRACTICE. 


11. 


1.   Multiply  7  ft.  3'  4''  5'"  by  5  ft.  8'    6'',  extending  the  pro- 
duct only  to  primes.  Ans,  41  ft.  7'it 
20 


4  ft. 

7' 

7" 

4  ft. 

3' 

11" 

3' 

8" 

6'" 

3' 

8" 

6'" 

230  DUODECIMALS. 

2.  IIow  many  yards  of  carpeting  will  cover  a  floor  86  ft.  9'  4" 
long,  and  26  ft.  6'  9''  wide  ? 

3.  How  many  cu.  ft.  in  a  block  of  marble  measuring  6  ft.  2'  7" 
in  length,  3  ft.  3'  4''  wide,  and  2  ft.  8'  6''  thick? 

4.  Find  the  product  of  7  ft.  6'  8'',  3  ft.  2'  IF,  and  3  ft.  8'  4", 
correct  to  within  1'.  Ans.  90  ft.  &db. 

DIVISION. 

S9f^.   I.  Divide  41  ft.  8'  7"  6'"  by  7  ft.  5'. 

OPERATION.  Analysis.       Divid- 

7  ft.  5')41  ft.  8'  7''  6'''(5  ft.  7'  6''  ing  the  units  of  the 

37  ft.  1'  dividend  by  the  units 

of  the  divisor,  we  ob- 
tain 5  ft.  for  the  first 
term  of  the  quotient, 
and  4  ft.  7^  fcr  a  re- 
mainder. Bringing 
down  the  next  term  of  the  dividend,  we  have  4  ft.  7^  7^^  for  a  new 
dividend.  Reducing  the  first  two  terms  to  primes,  we  have  55^  7^^, 
whence  by  trial  division  we  obtain  7^  for  the  second  term  of  the  quo- 
tient, and  y  8^^  for  a  remainder.  Completing  the  division  in  like 
manner,  we  have  5  ft.  7^  6^^  for  the  entire  quotient  Hence  the  fol- 
lowing 

Rule.  I.  Write  the  divisor  on  the  left  hand  of  the  dividend^ 
as  in  simple  numbers, 

II.  Find  the  first  term  of  the  quotient  either  hy  dividing  the 
first  term  of  the  dividend  hy  the  first  term  of  the  divisor ,  or  hy 
dividing  the  first  two  terms  of  the  dividend  hy  the  first  two  terms 
of  the  divisor  ;  multiply  the  divisor  hy  this  term  of  the  quotient, 
subtract  the  product  from  the  corresponding  terms  of  the  dividend y 
and  to  the  remainder  hring  down  another  term  of  the  dividend. 

III.  Proceed  in  like  manner  till  there  is  no  remainder,  or  till  a 
quotient  has  been  obtained  snfiiciently  exact. 

EXAMPLES    FOR   rRACTICE. 

1  Divide  287  ft.  T  by  17  ft.  Ans.  16  ft.  11'. 

2.  Divide  29  ft.  b'  V  by  6  ft.  8'.  Ayis.  4  ft.  5'. 


I 


DIVISION.  231 

3.  A  floor  whose  length  is  48  ft.  6'  has  an  area  of  1176  ft.  1' 
6'';  what  is  its  width?  Ans.  24  ft.  3'. 

4.  From  a  cellar  38  ft.  10'  long  and  9  ft.  4'  deep,  were  exca- 
vated 275  cu.  yd.  5  cu.  ft.  1'  4^'  of  earth;  how  wide  was  the 
cellar  ?  .  Ans.  20  ft.  6'. 

CONTRACTED    METHOD. 

3f^3.  Division  of  Duodecimals  may  be  abbreviated  after  the 
manner  of  contracted  division  of  decimals. 

1.  Divide  35  ft.  11'  11"  by  4  ft.  3'  7"  3'",  and  find  a  quotient 
correct  to  seconds. 

OPERATION". 

4  ft.  3'  7"  3'"  )  35  ft.  ir  11"  (  8  ft.  4'  5" 
34  ft.     4'  10" 


Ift. 

7' 

1" 

1ft. 

5' 

2" 

1' 

11" 

1' 

9" 

2",  rem. 

Analysis.  Having  obtained  by  trial,  8  ft.  for  the  first  term  of  the 
quotient,  we  multiply  three  terms  of  the  divisor,  4  ft.  y  7^^,  carrying 
from  the  rejected  term,  3^^^  X  8  ==  24^^^  =- 2^^  making  34  ft.  4'  W, 
which  subtracted  from  the  dividend  leaves  1  ft.  7^  1^^  for  a  new  divi- 
dend. In  the  next  division,  we  reject  2  terms  from  the  right  of  the 
divisor,  and  at  the  last  division,  3  terms,  and  obtain  for  the  required 
quotient,  8  ft.  4^  5^^. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  7  ft.  7'  3"  by  2  ft.  10'  7";  extending  the  quotient  to 
seconds.  Ans.  2  ft.  7'  8"=±=. 

2..  Separate  G4  ft.  9'  8"  into  three  factors,  the  first  and  second 
of  which  shall  be  7  ft.  2'  4"  and  4  ft.  7'  0"  8'"  respectively,  and 
obtain  the  ihird  factor  correct  to  within  1  second. 

Ans.   1  ft.  11'  3"d=. 

3.  What  is  the  width  of  a  room  whose  area  is  3G  ft.  4'  8"  and 
whose  lenj^th  7  ft.  2'  11"? 


232  SHORT  METHODS 

SHORT  METHODS. 
394.  Under  the  heads  of  Contractions  in  Multiplication  and 
Contractions  in  Division,  are  presented  only  such  short  nietliods 
as  are  of  the  most  extensive  application.  The  short  methods 
which  follow,  although  limited  in  their  application,  are  of  much 
value  in  computations. 

FOR  SUBTRACTION. 
393.   When  the   minuend  consists  of  one  or  more 
digits  of  any  order  higher  than  the  highest  order  in  the 
subtrahend. 

The  difference  between  any  number  and  a  unit  of  the  next 
higher  order  is  called  as  Arithmetical  Complement.  Thus,  4  is  the 
arithmetical  complement  of  6,  31  of  G9,  2792  of  7208,  etc. 

1.  Subtract  29876  from  400000. 

OPERATION.  Analysis.     To  subtract  29876  from  400000  is  the 

400000  same  as  to  subtract  a/number  one  less  than  29876,  or 

29876  29875,  from  399999.,  (Ax.  2).     We  therefore  diminish 

3701'^ 4  *^^  ^  of  the  minuend  by  1,  and  then  take  each  figure 

of  the  subtrahend  from  9,  except  the  last  or  right- 
hand  digit,  which  we  subtract  from  10.     Ilence  the 

HuLE.  I.  Subtract  \  from  the  njnificint  part  of  the  minuend 
and  write  the  remainder^  if  any ^  as  a  pctrt  of  the  result, 

IT.  Proceeding  to  the  right^  subtract  each  figure  in  the  subtra- 
hend from  9,  except  the  last  significant  figure^  which-  subtract 
from  10. 

EXAMPLES   FOR   PRACTICE. 

1.  Subtract  756  from  1000.  Ans,  244. 

2.  Subtract  8576  from  4000000.  Ans.  3991424. 

3.  Subtract  .5768  from  10. 

4.  Subtract  13057  from  1700000. 

5.  Subtract  90.59876  from  64000. 

6.  Subtract  599948  from  1000000. 

7.  What  is  the  arithmetical  complement  of  271  ?  Of  18365  1 
Of  3401250? 


FOR  MULTIPLICATION.  233 

FOR   MULTIPLICATION. 
CASE  L 

•I96.  When  the  multiplier  is  9,  99,  or  any  number 
of  9^s. 

Annexing  1  cipher  to  a  number  multiplies  it  by  10,  two  ciphers  by 
100,  three  ciphers  by  1000,  etc.  Since  9  is  10  —  1,  any  number  may 
be  multiplied  by  9  by  annexing  1  cipher  to  it  and  subtracting  the 
number  from  the  result.  For  similar  reasons,  100  times  a  number  — • 
1  time  tiie  number  =  99  times  the  number,  etc.     Hence, 

Rule.  Annex  to  the  multiplicand  as  many  ciphers  as  the  multi^ 
plier  contains  d^s,  and  subtract  the  multiplicand  from  the  result, 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  784  by  99.  Ans.  77616. 

2.  Multiply  5873  by  .999. 

3.  Multiply  4783  by  99999.  Ans.  478295217. 

4.  Multiply  75  by  999.999. 

CASE   II. 

397.  When  the  multiplier  is  a  number  a  few  units 
less  than  the  next  higher  unit. 

Were  we  required  to  multiply  by  97,  which  is  100  —  3,  we  could 
C3vidently  annex  2  ciphers  to  the  multiplicand,  and  subtract  3  times 
the  multiplicand  from  the  result.  Were  our  multiplier  991,  which  is 
1000  —  9,  we  could  subtract  9  times  the  multiplicand  from  1000  times 
the  multiplicand.     Hence, 

RuLSl.  I.  Mnltipli/  hy  the  next  higher  unit  hy  annexing 
ciphers. 

II.  From  this  result  siihtract  as  many  times  the  multiplicand 
as  there  are  units  in  the  difference  between  the  multiplier  and  the 
next  higher  unit, 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  786  by  98.  Ans.  77028. 

2.  Multiply  4327  by  96.  Ans.  415392. 

3.  Multiply  7328  by  997. 

20* 


234  BHOBT  METHODS 

4.  Multiply  7873.586  by  9.95.  Ans.  78342.18070. 

5.  Multiply  43789  by  9994. 

6.  Multiply  7077364  by  .999993. 

CASE   III. 

398.  "Wlieii  the  left  hand  figure  of  the  miiltiplier  is 
the  unit,  1,  the  right  hand  figure  is  any  digit  whatever, 
and  the  intervening  figures,  if  any,  are  ciphers. 

I.  Multiply  3684  by  17. 

oPERATiox.  Analysis.     If  we  multiply  by  the  usual 

3684  X   17  method,  we  obtain,  separately,  7  times  and 

nl)^^  10  times  the  multiplicand,  and  add  them. 

We  may  therefore  multiply  by  the  7  units, 

and  to  the  product  add  the  multiplicand  regarded  as  tens,  thus :  7  times 

4  is  28,  and  we  write  the  8  as  the  unit  figure  of  the  product.     Then, 

7  times  8  is  56,  and  the  2  reserved  being  added  is  58,  and  the  4  in 

the  multiplicand,  added,  is  G2,  and  we  write  2  in  the  product.     Next, 

7  times  6,  plus  the  6  reserved,  plus  the  8  in  the  multiplicand,  is  56, 

and  we  write  6  in  the  product.     Next,  7  times  3,  plus  the  5  reserved, 

plus  the  36  in  the  multiplicand,  is  62,  which  we  write  in  the  product, 

and  the  work  is  done. 

Had  the  multiplier  been  107,  we  should  have  multiplied  two  figures 
of  the  multiplicand  by  7,  before  we  commenced  adding  the  digits  of 
the  multiplicand  to  the  partial  products ;  3  figures  had  the  multiplier 
been  1007,  etc.     Hence  the 

Rule.  I.  Write  the  multiplier  at  the  right  of  the  multiplicand, 
with  the  sign  of  multiplication  between  them. 

II.  Multiply  the  multiplicand  hi/  the  unit  fijure  of  the  midti- 
plier^  cnid  to  the  product  add  the  midtiplicandy  regarding  its  local 
value  as  a  product  hy  the  left  hand  figure  of  the  multiplier, 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  567  by  13.  Ans.  7371. 

2.  Multiply  439603  by  10.5.  Ans.  4615831.5. 

3.  Multiply  7859  by  107. 

4.  Multiply  18075  by  1008.  Ans.  18219600. 

5.  Multiply  3907  by  10.002. 


FOR  MULTIPLICATION. 


235 


CASE    IV. 

tl99.  "When  the  left  hand  figure  of  the  multiplier  is 
any  digit,  the  right  hand  figure  is  the  unit,  1,  and  the 
intermediate  figures,  if  any,  are  ciphers. 

I.  Multiply  834267  by  301. 

OPERATION.  Analysis.     Regarding  the  multipli- 

83-l<267  X  301  cand  as  a  product  by  the  unit,  1,  of  the 

—7-  multiplier,  we    multiply  the   multipli- 

2511143G7  ^^^^  i^y  3  hundreds,  and  add  the  digits 

of  the  multiplicand  to  the  several  products  as  we  proceed.  Since  the 
3  is  hundreds,  the  two  right  hand  figures  of  the  multiplicand  will 
be  the  two  right  hand  figures  of  the  product ;  and  the  product  of  3x7 
will  be  increased  by  2,  the  hun:Ireds  of  the  multiplicand. 

Had  the  multiplier  been  31,  the  teiis  of  the  multiplicand  would 
have  been  added  to  3  X  7  ;  had  the  multiplier  been  3001  the  thousands 
of  the  multiplicand  would  have  been  added  to  3.  X  7 ;  and  so  on. 
Hence  the 

Rule.  I.  Write  the  multijpUer  at  the  right  of  the  multiplicand, 
with  the  sign  of  multiplication  hctwecn  them, 

II.  Multiply  the  multiplicand  hy  the  left  hand  figure  of  the  mid- 
tlpller^  and  to  the  product  add  the  multiplicand,  regarding  its  local 
value  as  a  product  hy  the  unit  figure  of  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  56783  by  71. 

2.  Multiply  47.89  by  60.1.  Ans.  2878.189. 

3.  Multiply  3724.5  by  .901 

4.  Multiply  103078  by  40001.  Ans.  4123223078. 

CASE   V. 

40@.  "When  the  digits  of  the  multiplier  are  all  the 
same  figure. 

1.  Multiply  81362  by  333. 

OPERATION.  Analysis.     AYe  first  multiply  by  999,  by 

81362000  (396).    Then,  since  333  is  \  of  999,  we  take 

81362  \  of  the  product. 

^  ^  o  i9Qppoo  Had  our  multiplier  been  444,  we  would 

have  taken  |  of  999  times  the  multiplicand. 

27093546  Had  it  been  GG,  we  would  have  taken  J  =  f 

of  99  times  the  multiplicand,  etc.     Ilenco 


236  SHORT  METHODS 

Rule  I.  Multiply  hy  as  many  9's  as  the  multiplier  contains 
diijlts,  hy  (396). 

II,  Take  siich  a  part  of  the  product  as  1  digit  of  the  multiplier 
is  part  of  9. 

EXAMPLE'S   FOR   PRACTICE. 

1.  Multiply  432711  by  222.  Ans.  9606184.2. 

2.  Multiply  578  by  1111. 

3.  Multiply  .6732  by  88.888.  Ans.  59.8394016. 

4.  Multiply  8675  by  77.7. 

5.  Multiply  44444  by  88888. 

CASE   VI. 

401.  To  square  a  number  consisting  of  only  two 
digits. 

I.  What  is  the  square  of  18  ? 
Analysis.     According  to  (86),  we  have 

182  =,  18  X  18 
Now  if  one  of  these  factors  be  diminished  by  2,  the  product  will  be 
less  than  the  square  of  18  by  2  times  the  other  factor,  (93,  I) ;  that  is, 
182=  (16  X  18)  4- (2  X  18). 
Next,  if  we  increase  the  other  factor,  18,  in  this  result,  by  2,  the 
whole  result  will  exceed  the  square  of  18,  by  2  times  the  other  factor, 
16,  (93,  III);  that  is, 

18-^  =  (16  X  20)  +  (2  X  18)  —  (2  X  IG). 

But  as  2  times  18  minus  2  times  16  is  equal  to  2x2,  or  2^,  we 

have 

182  =  16  X  20  +  22.     Hence  the 

Rule.  I.  Take  two  numbers,  one  of  ichich  is  as  many  units 
less  than  the  number  to  be  squared  as  the  other  is  units  greater j  and 
one  of  the  numbers  talcen  an  exact  number  of  tens. 

II.  Multiply  these  two  numbers  together j  and  to  the  product  add 
the  square  of  the  difference  between  the  given  and  one  of  the  as- 
sumed numbers. 

Note.  —  A  little  practice  will  enable  the  pupil  to  readily  square  any  number 
less  than  100  mentally  by  this  rule. 


f 


p 


FOR  MULTIPLICATION.  237 

EXAMPLES    FOR   PRACTICE. 

1.  ^^hat  is  the  square  of  27  ?  Ans.  729. 

2.  What  is  the  square  of  49  ?  Ajis.  2401. 

3.  Square  28,  26,  39,  38,  37,  36,  and  35. 

4.  Square  77,  88,  8.6,  99,  98,  69,  68,  6.7,  and  62. 

CASE   VII. 

40^.  When  the  multiplier  is  an  aliquot  part  of  some 
higher  unit. 

An  Aliquot  or  Even  Part  of  a  number  is  such  a  part  as  will 

exactly  divide  that  number.     Thus,  5,  8 J,  and  12  J  are  aliquot 

parts  of  25  and  of  100,  etc. 

Note. — An  aliquot  part  may  be  either  a  whole  or  a  mixed  number,  while  a 
com poutnt  factor  must  be  a  whole  number. 

403,    The  aliquot  parts  of  10  are  5,  3|,  2^,  2,  If,  If,  1{,  l\. 

The  aliquot  parts  of  100,  1000,  or  of  any  other  number,  may 
be  found  by  dividing  the  number  by  2,  3,  4,  etc.,  until  it  has 
been  divided  by  all  the  integral  numbers  between  1  and  itself. 

I.  Multiply  78  by  3i,  and  by  25  sepa^-ately. 

OPERATION.  Analysis.     Since  3J  is  J  of  10, 

3  )  780  4  )  7800  the  next  higher  unit,  we  multiply 

~260    ^  1950  "^^  ^y  ^^  ^^^  ^^^®  ^  ^^  ^^^  product. 

Again,  since  25   is  }  of  100,  we 
multiply  78  by  100  and  take  J  of  the  product.     Hence  the 

Rule.  I.  MultipJi/  the  given  multiplicand  hij  the  unit  next 
higher  thfin  the  multiplier^  hy  annexing  ciphers. 

II.  T'ake  such  a  part  of  this  product  as  the  given  multiplier  is 
part  of  the  next  higher  unit. 

EXAiMPLES    FOR   PRACTICE. 

1.  Multiply  437  by  25.  Ans.  10925. 

2.  Multiply  6872  by  2-|.  Ans.  17180. 

3.  Multiply  5734154  by  333^.  Ans.  1911384666|. 

4.  Multiply  758642  by  12-^. 

5.  Multiply  78563  by  125.  Ans.  9820375. 

6.  Multiply  57687  by  142f 


238 


SHORT  METHODS 


CASE  viir. 
404.   AYlien  the  right  hand  figure  or  figures  of  the 
multiplier  are  aliquot  parts  of  10,  100,  1000,  etc. 
1.  Multiply  2183  by  1233i. 

OPERATION. 


218300 
12^ 

727661 
26196 


Analysis.  1233J  =  12J  X  100.  We  there- 
fore multiply  by  100,  and  by  12J,  in  continued 
multiplication.     Hence  the 


26923661 

Rule.  I.  Reject  from  the  right  hand  of  the  multiplier  such 
fgnre  or  fgures  as  are  an  aliquot  j^cirt  of  some  higher  uuit,  and 
to  the  remaining  figures  of  the  m^ultiplier  annex  a  fraction  v:hich 
expresses  the  aliquot  part  thus  rejected ^  for  a  reserved  multiplier. 

II.  Annex  to  the  multiplicand  as  mani/  ciphers  as  are  equal  to 
the  number  of  figures  rejected  from  the  right  hand  of  the  multi- 
plicTy  and  multiple/  the  result  hy  the  reserved  multiplier. 


EXAMPLES    FOR   PRACTICE. 


1.  Multiply  43789  by  825. 

2.  Multiply  58730  by  7125. 

3.  Multiply  7854  by  34.2}. 

4.  Multiply  30724  by  73333i 

5.  Multiply  47836  by  712}. 

6.  Multiply  53727  by  2416f . 


Ans.  36125925. 


Ans.  268999.5. 


Ans.  34083150. 


CASE    IX. 

4©^.   To  find  the  cost  of  a  quantity  when  the  price 
is  an  aliquot  part  of  a  dollar. 

1.  What  cost  a  case  of  muslins  containing  1627  yds.,  at  ^.12} 
per  yard  ? 

OPERATION.  Analysts.    At  $1  per  yard  the  case  would 

8  )  $1627  cost  as  many  dollars  as  it  contained  yards ; 

g.9Qo  0^1  ai^d  at  $.12}  =  $4  per  yard,  it  would  cost  } 

as  many  dollars  as  it  contained  yards.     Wo 

therefore  regard  the  yards  as  dollars,  which  we  divide  by  8.     Hence, 


R 


FOR  MULTIPLICATION.  239 

KuLE.      Tahe  such  a  part  of  the  given  quantitij  as  the  price  is 

part  of  one  dollar. 

NoTK.  —  Since  the  shilling  in  most  of  the  different  currencies  is  some  aliquot 
part  of  the  dollar,  this  rule  is  of  much  practical  use  in  making  out  bills  and 
accounts  where  the  prices  of  the  items  are  given  in  State  Currency,  and  the 
amounts  are  required  in  United  States  Money. 

EXAMPLES    FOR   PRACTICE. 

1.  What  cost  5G8  pounds  of  butter  at  25  cents  a  pound  ? 

Am,  S142. 

2.  A  merchant  sold  51  yards  of  prints  at  16f  cents  per  yard,  8 
pieces  of  sheeting,  each  piece  containing  83  yards,  at  6}  cents  per 
yard,  and  received  in  payment  18  bushels  of  oats  at  33i  cents  per 
bushel,  and  the  balance  in  money ;  how  much  money  did  he  re- 
ceive ?  Ans.   $19. 

3.  Required  the  cost  of  28  dozen  candles,  at  1  shilling  pei 
dozen.  New  York  currency.  Ans.  $3.50. 

4.  What  cost  576  lbs.  of  beef  at  lOd.  per  pound,  Pennsylvania 
currency?  Ans.  §64. 

5.  If  a  grocer  in  New  York  gain  §7.875  on  a  hogshead  of  mo^ 
lasses  containing  63  gallons,  how  much  will  he  gain  on  576  gallons 
at  the  same  rate  ?  Ans.  §72. 

CASE    X. 

406.  To  find  the  cost  of  a  quantity,  when  the  quan- 
tity is  a  compound  number,  some  part  or  all  of  which  is 
an  aliquot  part  of  the  unit  of  price. 

1.  What  cost  5  bu.  3  pk.  4  qt.  of  cloversced,  at  §3.50  per  bu.  ? 

OPERATION.  Analysis.     Mnltiplj- 

2„   4,,  8  )  §3.50  price.  ing  the  price  by  5,  ^vo 

5  have   the  cost  of  5    bu. 

SrTsO    cost  of  5  bu.  Dividing  the  price  by  2, 

1  75        "     "  2  nk  ^^  have  the  cost  of  \  bu. 

.875      u      a   -^     u  =2   pk.     Dividing   the 

.4375   ^'      ^^   4  qt.  price  by  4,  or  the  cost  of 

ooTrkTw"     A  2  pk.  by  2,  Ave  have  the 

$20.06.: 5,  ^/? 5.  \     /     '  . 

^  cost  of  1   pk.     Dividing 

the  price  by  8,  or  the  cost  of  2  pk.  by  4,  or  the  cost  of  1  pk.  by  2,  we 


240  SHORT  METHODS 

have  the  cost  of  J  pk.  ===  4  qt.     And  the  sum  of  these  several  values 
is  the  entire  cost  required. 

2.   At  £6  7s.  5 id.  Sterling  per  hhd.,  how  much  will  4  hhd.  9 
gal.  3  qt.  of  West  India  Molasses  cost  ? 

OPERATION.  Analysis.     Mul- 

7)  £6      7s.       5  Jd.  price.  tiplying    the    price 

4  by  4,  we  have  the 

"  25     9  ''    10  "         cost  of  4  hhd.  ^^^t  of  4  hhd.    Di- 

12)  ''          18  "      2   "  2  qf. ''    "  9  gal.  viding  the  price  by 

1  "      6   ^^      I  ^^  "    ^^  3  qt.  7,  we  have  the  cost 

"  26     9  *^      6  ''  2|,  "  Ans,  ^{  \  ^^^-  ="  ^  S^^- 

Dividing  the  cost  of 
9  gal.  by  12,  we  have  the  cost  of  j^2  of  36  qt.  =  3  qt.     And  the  sum 
of  these  several  results  is  the  entire  cost  required. 
From  these  illustrations  we  deduce  the  following 
Rule.     I.  Mullipli/  the  price  hy  the  number  of  units  of  the  de- 
nomination corresponding  to  the  price. 

II.  For  the  lower  denominations^  take  aliquot  parts  of  the  price; 
the  sum  of  the  several  results  will  he  the  entire  cost. 

Note. — This  method  is  applicable  in  certain  cases  of  multiplication,  where  one 
compound  number  is  taken  as  many  times  as  there  are  units  and  parts  of  a 
unit  of  a  certain  kind,  in  another  compound  number.  This  will  be  $een  in  the 
first  example  below. 

EXAMPLES   FOR   PRACTICE. 

1.  A  chemist  filtered  18  gal.  3  qt.  1  pt.  of  rain-water  in  1  day; 

at  the  same  rate  how  much  could  he  filter  in  4  da.  6  h.  30  min.  ? 

OPERATION.  Analysis.     Multi- 

18  gal.  3  qt.   1  pt.  in  1  da.  VV^^    the    quantity 

4  filtered  in  1  day  by  4, 

,^     .     ^^  we  have  the  quantity 

^i    />  1  filtered  in  4  days.  Di- 

^"    "  30  min.  viding  the  quantity  fil- 

-  tered  in  1  day  by  4, 

80    "     2  "    0  "  3  Jj  "     Ans.  ^e  have  the  quantity 

filtered  in  }  da.  =  6 
h.  Dividing  the  quantity  filtered  in  6  hours  by  12,  we  have  the  quan- 
tity filtered  in  J  h.  =  30  min.  And  the  sum  of  these  several  results 
is  the  entire  result  required. 


75 

U 

2  '^ 

0  " 

4 

iC 

2  '^ 

1  "  3 

1  ^^ 

la        7 

FOR  DrV'ISION. 


241 


I 


2.  What  will  be  the  cost  of  3  lb.  10  oz.  8  pwt.  5}  gr.  of  gold 
at  $15.46  per  oz.  ?  Aiis.  $717.52. 

3.  A  man  bought  5  cwt.  90  lb.  of  hay  at  $.56  per  cwt. ;  what 
was  the  cost?  Ans.  $3,304. 

4.  What  must  be  given  for  3  bu.  1  pk.  3  qt.  of  cloverseed,  at 
$4.48  per  bushel?  Jns.  $14.98. 

5.  A  gallon  of  distilled  water  weighs  8  lb.  5  oz.  6.74  dr. ;  re- 
quired the  weight  of  5  gal.  3  qt.  1  pt.  3  gi. 

Ans.  49  lb.  12  oz.  5.73—  dr. 

6.  At  $17.50  an  acre,  what  will  3  A.  1  R.  35.4  P.  of  land  cost? 

7.  If  an  ounce  of  English  standard  gold  be  worth  £3  17s.  lOH-j 
what  will  be  the  value  of  an  ingot  weighing  7  oz.  16  pwt.  18  gr.  ? 

Ans.  £30  10s.  4.14375d. 

8.  If  a  comet  move  through  an  arc  of  4°  36'  40"  in  1  day,  how 
far  will  it  move  in  5  da.  15  h.  32  min.  55  sec.  ? 

9.  What  will  be  the  cost  of  7  gal.  1  qt.  1  pt.  3  gi.  of  burning 
fluid,  at  4s.  8d.  per  gallon,  N.  Y.  currency?         Am.  $4.35  +  . 

10.  What  must  be  paid  for  12  J  days'  labor^  at  5s.  3d.  per  day, 
New  England  currency  ? 

FOR  DIVISION. 
CASE   I. 

407.  "When  tlie  divisor  is  an  aliquot  part  of  some 
higher  unit. 

1.  Divide  260  by  3 J,  and  1950  by  25. 

OPERATION.  Analysis.     Since  3J  is  J  of  10,  the  next 

26|0  19|50  higher  unit,  we  divide  260  by  10 ;  and  hav- 

3     and    4  ing  used  3  times  our  true  divisor,  we  obtain 

no  Wo  only  i  of  our  true  quotient.      Multiplying 

the  result,  26,  by  3,  we  have  78,  the  true 

quotient.    Again,  since  25  is  }  of  100,  the  next  higher  unit,  we  divide 

1950  by  100 ;  and  having  used  4  times  our  true  divisor,  the  result, 

19.5,  is  only  }  of  our  true  quotient.     Multiplying  19.5  by  4,  we  have 

78,  the  true  quotient.     Hence  the 

Rule.     I.   Divide  the  given  dividend  hy  a  unit  of  the  order 
next  higher  than  the  divisor y  hy  cutting  off  ^qures  from  the  right. 
21  Q 


242 


SHOET  METHODS. 


II.    Talce  as  manr/  times  this  quotient  as  the  divisor  is  contained 
times  in  the  next  higher  unit. 


EXAMPLES   FOR   PRACTICE. 

1.  Divide  63475  by  25. 

2.  Divide  7856  by  1.25. 

3.  Divide  516  by  33.3|. 

4.  Divide  16.7324  by  12J. 

5.  Divide  1748  by  .14f. 

6.  Divide  576.34  by  1.6f. 


Ans.  2539. 
Alls.  6284,S. 


Ans.  12236 


OPERATION. 


CASE    IT. 

408.  When  the  right  hand  figure  or  figures  of  the 
divisor  are  an  aliquot  part  of  10,  100,  1000,  etc. 
1.  Divide  26923661  by  1233J. 

Analysis.  Since  33J  is 
J  of  100,  we  multiply  both 
dividend  and  divisor  by  3, 
(117,  HI),  and  we  obtain  a 
divisor  the  component  fac- 
tors of  which  are  100  and  37. 
"VYe  then  divide  after  the 
manner  of  contracted  divi- 
sion, (112). 


1233J)  26923661 
3_ 3^ 

37|00)  80771100  (2183,^715. 
67 
807 
111 

.  Divide  601387  by  1875. 

OPERATION. 


Analysis.  Multiplying  both 
dividend  and  divisor  by  4,  we  ob- 
tain a  new  divisor,  7500,  having  2 
ciphers  on  the  right  of  it.  Multi- 
plying again  by  4,  we  obtain  a  new 
divisor,  30000,  having  4  ciphers  on 
the  right.  Then  dividing  the  new 
dividend  by  the  new  divisor,  we  ob- 
tain 320  for  a  quotient,  and  22192 
for  a  remainder.  As  this  remainder  is  a  part  of  the  new  dividend, 
it  must  be  4  X  4  =  16  times  the  true  remainder ;  we  therefore  divide 
it  by  16,  and  write  the  result  over  the  given  divisor,  1875,  and  annex 
the  fraction  thus  formed  to  the  integers  of  the  quotient. 


1875)   601387 
4 4 

7500  )  2405548 
4   4 

310000)  96212192 


320i|||,  Ans, 


RATIO.  243 

Erom  these  illustrations  we  derive  the  following: 

E-ULE.      I.   Multiply  both  dividend  and  divisor  hj  a  number  or 

numbers  that  will  produce  for  a  new  divisor  a  number  ending  in  a 

cip)her  or  ciphers. 

II.   Divide  the  new  dividend  by  the  new  divisor. 

Note. — If  the  divisor  be  a  whole  number,  or  a  finite  decimnl,  the  multiplier 
will  be  2,  4,  5,  or  8,  or  some  multiple  of  one  of  these  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  64375  by  2575. 

2.  Divide  76394  by  3625.  Ans.  2^%%%. 
8.  Divide  7325  by  433J. 

4.  Divide  5736  by  431.25.  Ans.  l^^. 

5.  Divide  42.75  by  566f. 

6.  Divide  24409375  by  .21875. 

7.  Divide  785  by  3.14f.  .   Ans.  249^|. 


RATIO. 

4L09.   Ratio  is  the  relation  of  two  like  numbers  with  respect 

to  comparative  value. 

Note.  —  There  are  two  methods  of  comparing  numbers  with  respect  to  value; 
1st,  by  subtracting  one  from  the  other;  2d,  by  dividing  one  by  the  other.  The 
relation  expressed  by  the  difference  is  sometimes  called  An'thnu'tical  Jiati'o,  and 
the  relation  expressed  by  the  quotient,  Geometrical  Itutio. 

410.  When  one  number  is  compared  with  another,  as  4  with 
12,  by  means  of  division,  thus,  12  -^4  =  3,  the  quotient,  3,  shows 
the  relative  value  of  the  dividend  when  the  divisor  is  considered 
as  a  unit  or  standard.  The  ratio  in  this  case  shows  that  12  is  3 
times  4 ;  that  is,  if  4  be  regarded  as  a  unit,  12  will  be  3  units,  or 
the  relation  of  4  to  12  is  that  of  1  to  3. 

4:11.    Ratio  is  indicated  in  two  ways : 

1st.  By  placing  two  points  between  the  two  numbers  compared, 
writing  the  divisor  before  and  the  dividend  after  the  points. 
Thus,  the  ratio  of  8  to  24  is  written  8  :  24;  the  ratio  of  7  to  5  is 
written  7  :  5. 


244  RATIO. 

2d.  In  the  form  of  a  fraction.  Thus,  the  ratio  of  8  to  24  is 
written  \^  -,  the  ratio  of  7  to  5  is  ^. 

4: IS.    The  Terms  of  a  ratio  are  the  two  numbers  compared. 

The  Antecedent  is  the  first  term;  and 

The  Consequent  is  the  second  term. 

The  two  terms  of  a  ratio  taken  together  are  called  a  couplet, 

48 S,   A  Simple  Ratio  consists  of  a  single  couplet;  as  5  :  15. 

414.  A  Compound  Eatio  is  the  product  of  two  or  more  sim- 
ple ratios.     Thus,  from  the  two  simple  ratios,  5  :  16  and  8  :  2,  we 

5  :  16 

82_2 

may  form  the  compound  ratio  5x8  :  16x 2,  or  ^^^  X  §  =  |§  =  |. 

41«5.  The  Reciprocal  of  a  ratio  is  1  divided  by  the  ratio ;  or, 
which  is  the  same  thing,  it  is  the  antecedent  divided  by  the  con- 
ficquent.  Thus,  the  ratio- of  7  to  9  is  7  :  9  or  |,  and  its  reciprocal 
is  |. 

Note.  —  The  quotient  of  the  second  term  divided  by  the  first  is  sometimes 
called  n  Direct  liatio,  and  the  quotient  of  the  first  term  divided  by  the  second, 
jin  Inverse  or  Reciprocal  Ilatio. 

4115.  One  quantity  is  said  to  vary  directly  as  another,  when 
the  two  increase  or  decrease  together  in  the  same  ratio ;  and  one 
quantity  is  said  to  vary  inversely  as  another,  when  one  increases 
in  the  same  ratio  as  the  other  decreases.  Thus  time  varies  directly 
as  wages ;  that  is,  the  greater  the  time  the  greater  the  wages,  and 
the  less  the  time  the  less  the  wages.  Again,  velocity  varies  in- 
versely  as  the  time,  the  distance  being  fixed;  that  is,  in  traveling 
a  given  distance,  the  greater  the  velocity  the  less  the  time,  and  the 
less  the  velocity  the  greater  the  time. 

41*?.  Ratio  can  exist  only  between  like  nunabers,  or  between 
two  quantities  of  the  same  kind.  But  of  two  unlike  numbers  or 
quantities,  one  may  vary  either  directly  or  inversely  as  the  other. 
Thus,  cost  varies  directly  as  quantity,  in  the  purchase  of  goods; 
time  varies  inversely  as  velocity,  in  the  descent  of  falling  bodies. 
In  all  cases  of  this  kind,  the  quantities,  though  unlike  in  kind, 
have  a  mutual  dependence,  or  sustain  to  each  other  the  relation 
of  cause  and  effect. 


RATIO.  245 

4:18.    In  the  comparison  of  like  numbers  we  observe, 

I.  If  the  numbers  are  simple^  whether  abstract  or  concrete, 
their  ratio  may  be  found  directly  by  division. 

II.  If  the  numbers  are  componndj  they  must  first  be  reduced 
to  the  same  unit  or  denomination. 

III.  If  the  numbers  are  fractional,  and  have  a  common  de- 
nominator, the  fractions  will  be  to  each  other  as  their  numerators; 
if  they  have  not  a  common  denominator,  their  ratio  may  be  found 
either  directly  by  division,  or  by  reducing  them  to  a  common 
denominator  and  comparing  their  numerators. 

4H9.  Since  the  antecedent  is  a  divisor  and  the  consequent  a 
dividend,  any  change  in  either  or  both  terms  will  be  governed  by 
the  general  principles  of  division,  (117).  AVe  have  only  to  sub- 
stitute the  terms  antecedent^  consequent^  and  rat'iOy  for  dluuor, 
dividend^  and  quotient^  and  these  principles  become 

GENERAL    PRINCIPLES    OF   RATIO. 

Prin.  I.  Multiplying  the  consequent  multiplies  the  ratio ;  divi- 
ding the  consequent  divides  the  ratio. 

Prix.  II.  Multiplying  the  antecedent  divides  the  ratio  ;  dividing 
the  antecedent  multiplies  the  ratio. 

Prin.  III.  Multiplying  or  dividing  hoth  antecedent  and  conse^ 
quent  hy  the  same  numhcr  does  not  alter  the  ratio. 

4^0.    These  three  princij)les  may  be  embraced  in  one 

GENERAL    LAW. 

A  change  in  the  consequent  hy  multiplication  or  division  produr 
ces  a  LIKE  change  in  the  ratio  ;  hut  a  change  in  the  antecedent 
produces  an  OPPOSITE  change  in  the  ratio. 

421.  Since  the  ratifO  of  two  numbers  is  equal  to  the  conse- 
quent divided  by  the  antecedent,  it  follows,  that 

I.  The  antecedent  is  equal  to  the  consequent  divided  by  the 
ratio;  and  that, 

II.  The  consequent  is  equal  to  the  antecedent  multiplied  by  the 
ratio. 

21* 


246  RATIO. 

EXAMPLES    FOR   PRACTICE. 

1.  What  part  of  28  is  7  '/ 

jTg  —  1 ;  or,  £8  :  7  as  1  :  i ;  that  is,  28  has  the  same  ratio  to  7  that 

1  has  to  }.  Ans.   J. 

2.  What  part  of  42  is  6? 

3.  What  is  the  ratio  of  120  to  80  ?  Ans,  |. 

4.  What  is  the  ratio  of  8^  to  60  ?  Ans,  7. 

5.  What  is  the  ratio  of  ^^3  to  26  ? 

6.  What  is  the  ratio  of  7^  to  2^?  Ans.  f?. 

7.  What  is  the  ratio  of  J  to  -j^^  ?  A71S.  44. 

8.  What  is  the  ratio  of  1  mi.  to  3  fur.  ?  Ans.  |. 

9.  What  is  the  ratio  of  1  wk.  3  da.  12  h.  to  9  wk.  ?  Ans.  6. 

10.  What  is  the  ratio  of  10  A.  1  E.  20  P.  to  6  A.  2  11.  30  P.  ? 

11.  What  is  the  ratio  of  25  bu.  2  pk.  6  qt.  to  40  bu.  4.5  pk.  ? 

12.  What  is  the  ratio  of  181°  to  45'  30''  ? 

l^^i      2  of  3 

13.  What  part  of  -^  is  ^—^  ?  ^  Ans.  ^f  3. 

113  93 

14.  What  is  the  ratio  of  — /  to  |  of  A  of  -^  ?     Ans.   q%%%. 

■   b^  i'6 

15.  Find  the  reciprocal  of  the  ratio  of  42  to  28.        Ans.  1^. 
IG.  Find  the  reciprocal  of  the  ratio  of  3  qt.  to  43  gal. 

17.  If  the  antecedent  be  15  and  the  ratio  |,  what  is  the  conse- 
quent? Ans.  12. 

18.  If  the  consequent  be  3|  and  the  ratio  7,  what  is  the  ante- 
cedent? A71S.   1|. 

19.  If  the  antecedent  be  ^  of  |  and  the  consequent  .75^  what 
is  the  ratio  ? 

20.  If  the  consequent  be  ^G.12J  and  the  ratio  25,  what  is  the 
antecedent?  Ans.  $.245. 

21.  If  the  ratio  be  J  and  the  antecedent  |,  what  is  the  conse- 
quent ? 

22.  If  the  antecedent  be  13  A.  3  11.  25  P.  and  the  ratio  ||, 
what  is  the  consequent  ?  Aiis.  6  A.  2  R.  10  P. 


PROPERTIES  OF  PROPORTION.  247 


PEOPORTION. 

4:22.  Proportion  is  an  equality  of  ratios.  Thus,  the  ratios 
5  :  10  arid  6  :  12,  each  being  equal  to  2,  form  a  proportion. 

Note.  —  When  four  numbers  form  a  proportion,  they  are  said  to  be  propor- 
tional. 

4:23.    Proportion  is  indicated  in  three  ways : 

1st.  By  a  double  colon  placed  between  the  two  ratios;  thus, 
3  : 4  :  :  9  :  12  expresses  the  proportion  between  the  numbers  3,  4, 
9,  and  12,  and  is  read,  3  is  to  4  as  9  is  to  12. 

2d.  By  the  sign  of  equality  placed  between  two  ratios;  thus, 
3  :  4  =  9  :  12  expresses  proportion,  and  may  be  read  as  above,  or, 
the  ratio  of  3  to  4  equals  the  ratio  of  9  to  12. 

3d.  By  employing  the  second  method  of  indicating  ratio;  thus, 
I  =  ^-^  indicates  proportion,  and  may  be  read  as  either  of  the 
above  forms. 

424.  Since  each  ratio  consists  of  two  terms,  every  proportion 
must  consist  of  at  least  four  terms.     Of  these 

The  Extremes  are  the  first  and  fourth  terms ;  and 

The  Means  are  the  second  and  third  terms. 

42«5.  Three  numbers  are  proportional  when  the  first  is  to  the 
second  as  the  second  is  to  the  third.  Thus,  the  numbers  4,  G, 
and  9  are  proportional,  since  4:6  =  6:9,  the  ratio  of  each  couplet 
being  |,  or  IJ. 

420.  When  three  numbers  are  proportional,  the  second  term 
is  called  the  Mean  Proportional  between  the  other  two. 

^*'^lf.    if  v,'o  Iiavc  any  proportion,  as 
3  :  15  =r  4  :  20, 

Then,  indicating  this  ratio  by  the  second  method,  we  have 

V  =  ?• 

Reducing  these  fractions  to  a  common  denominator, 
15  X  4  _  20  X  3 
~12  12     • 

And  since  these  two  equal  fractions  have  the  same  denominator, 
the  numerator  of  the  first,  which  is  the  product  of  the  means,  must 
be  equal  to  the  numerator  of  the  second,  which  is  the  product  of  the 
extremes  ;  or,  15  X  4  ^  20  X  3.     Hence, 


248  PROPORTION. 

I.  In  every  proportion  the  product  of  the  means  equals  the 
product  of  the  extremes. 

4.gain,  take  any  three  terms  in  proportion,  as 

4  :  6=6  :  9 
Then,  since  the  product  of  the  means  equals  the  product  of  the  ex- 
tremes, 

62  =  4  X  9.     Hence, 

II.  The  square  of  a  mean  proportional  is  equal  to  the  product 
of  the  other  two  terms. 

428.  Since  in  every  proportion  the  product  of  the  means 
equals  the  product  of  the  extremes,  (^rSy,  I),  it  follows  that,  any 
three  terms  of  a  proportion  being  given,  the  fourth  may  be  found 
by  the  following 

Rule.  I.  Divide  the  product  of  the  extremes  hy  one  of  the 
meanSj  and  the  quotient  will  be  the  other  mean.     Or, 

II.  Divide  the  product  of  the  means  hy  one  of  the  extremes^  and 
the  quotient  will  he  the  otlier  extreme. 

EXAMPLES    FOR   PRACTICE. 

The  required  term  in  an  operation  will  be  denoted  by  (?), 
which  may  be  read  "  how  many,"  or  "  how  much." 

Find  the  term  not  given  in  each  of  the  following  proportions : 

1.  4  :  26  =  10  :  (  ?  ).  Ans.  65. 

2.  $8865  :  ?720  =  (?)  :  16  A.  Ans,   197  A. 

3.  4|  yd.  :  (  ?  )  :  :  ?9.75  :  829.25.         Ans.   13^  yd. 

4.  (?)  :  21  A.  3  R.  20  P.  :  :  $1260  :  $750. 

Ans.  36  A.  3  R. 
6.     7.50:18  =  (?):7yVoz. 

6.  7  oz  :(?)::  £30  :  £407  2s.  lOf  d.        Ans.  7  lb.  11  oz. 

7.  (  ?  )  :  .15  hhd.  :  :  $2.39  :  $.3585.  Ans.  1  hhd. 

8.  1  T.  7  cwt.  3  qr.  20  lb.  :  13  T.  5  cwt.  2  qr.  =  $9.50  :  (  ?  ). 

9.  $175.35  :  (?)  =  I  :  f  Ans.  $601.20. 

10.  (?)  :  $12^  =  2404  •  149VAV  Ans.  $20|. 

11.  I  yd.  :(?)::  $1  :  $59.0625.  Ans.  40^  yd. 


SIMPLE  PROPORTION.  Ojn 

CAUSE    AND    EFFECT. 

^QQ»  Every  question  in  proportion  may  be  considered  as  a 
comparison  of  two  cau;es  and  two  cfftrti^.  Thus,  if  3  dollars  as 
V  came  will  buy  12  pounds  as  an  rffecty  G  dollars  as  a  can^e  wi'.l 
Duy  24  pounds  as  an  rffect.  Or,  if  5  horses  as  a  cause  consume 
10  tons  as  an  effect,  15  horses  as  a  cause  will  consume  oO  tons  as 
an  ("ffect. 

Causes  and  effects  in  proportion  are  of  two  kinds  —  simple  and 
compound. 

4*1^.  A  Simple  Cause  or  Effect  contains  but  one  clement; 
as  price,  quantity,  cost,  time^  distance,  or  any  single  factor  used 
as  a  term  in  proportion. 

431.  A  Compound  Cause  or  Effect  is  the  product  of  two  or 
more  elements;  as  the  number  of  workmen  taken  in  connection 
with  the  time  employed,  length  taken  in  connection  with  breadth 
and  depth,  capital  considered  with  reference  to  the  time  em- 
ployed, etc. 

43 ^«  Since  liJce  causes  will  always  be  connected  with  like 
effects^  every  question  in  proportion  must  give  one  of  the  following 
statements  : 

1st  Cause     :     2d  Cause     =     1st  Effect     :     2d  Effect. 
1st  Effect     :     2d  Effect     -^     1st  Cause     :     2d  Cause, 
in  which   the  two  causes  or   the  two  effects  forming  one  couplet, 
must  be  like  numbers  and  of  the  same  denomination. 

Considering  all  the  terms  of  a  proportion  as  abstract  numbers, 
we  may  say  that 

1st  Cause     :     1st  Effect     =     2d  Cause     :     2d  Effect. 

But  as  ratio  is  the  result  of  comparing  two  numbers  or  things 
of  the  same  hbid^  (4t'7);  the  first  form  is  regarded  as  the  more 
natural  and  philosophical. 

SIMPLE   PROPORTION. 

433.  Simple  Proportion  is  an  equality  of  two  simple  ratios, 
and  consists  of  four  terms. 

Questions  in  simple  proportion  involve  only  simple  causes  and 
simple  effects. 


250 


PROPOETION. 


STATEMENT. 

$  $  yds. 

8      :     12     =     86     ; 

1st  cause.    2d  cause.         1st  effect. 


8  X 


OPERATION. 

(?)=    12  X  36 

^4^  X  -i^p 


(?) 


^ 


yds. 

(?) 

2d  effect. 


=  54  yd. 


FIRST    METHOD. 

1.  If  S8  will  buy  86  yards  of  velvet,  how  many  yards  may  be 

bought  for  §12?  , 

Analysis.  The  re- 
quh*ed  term  in  this  ex- 
ample is  an  effect ;  and 
the  statement  is,  $8  is 
to  $12  as  3G  yards  is 
to  ( ?  ),  or  how  many 
yards.  Dividing  12  X 
3G,  the  product  of  the 
means,  by  8,  the  given 
extreme,  we  have  (  ?  ) 
=  54  yards,  the  re- 
quired term,  (428,  II). 

2.  If  6  horses  will  draw  10  tons,  how  many  horses  will  draw 
15  tons  ? 

Analysis.  In  this  ex- 
ample a  cause  is  required  ; 
and  the  statement  is,  6 
horses  is  to  ( ? ),  or  how 
many  horses,  as  10  tons  is 
to  15  tons.  Dividing  15  X 
6,  the  product  of  the  ex- 
tremes, by  10,  the  given 
mean,  we  have  9,  the  re- 
quired term,  (428,  I). 


horses. 

6     : 

1st  cause. 


STATEMENT. 

horses.  tons. 

(?)     =     10 

2d  cause.  1st  effect. 


tons. 

15 

2d  effect. 


OPERATION. 

3 


40 
(?) 


r 


9  horses. 

4:S4.    Hence  the  following 

IluLE.  I.  Arrange,  the  terms  in  the  statement  so  that  the  causes 
shall  compose  one  covp)let,  and  the  effects  the  other,  putting  (?)  in 
the  place  of  the  required  term. 

II.  If  the  required  term  he  an  extreme,  divide  tlie  product  of  the 
means  hy  the  given  extreme;  if  tlie  required  term,  he  a  mean, 
divide  the  product  of  the  extremes  hy  the  given  mean. 

NoTRS. — 1.  If  the  terms  of  nny  eotiplet  be  of  different  denoriiinations,  they 
mu8t  he  reduced  to  the  same  unit  value. 

2.  If  the  odd  term  be  a  compound  number,  it  must  be  reduced  either  to  its 
lowe.«t  unit,  or  to  a  fraction  or  a  decimal  of  its  highest  unit. 

?>.  If  the  divisor  and  dividend  contain  one  or  more  factors  common  to  both, 
they  should  be  canceled.     If  any  of  the  terms  of  a  proportion  contain  mixed 


SIMPLE^PROPOHTION.  251 

number?,  they  should  first  be  changed  to  improper  fractions,  or  the  fractional 
part  to  a  decimal. 

4.  When  the  vertical  line  is  used,  the  divisor  and  (?)  are  written  on  the  left, 
and  the  fiictors  of  the  dividend  on  the  right. 


SECOND    METHOD. 

^3^.  The  following  method  of  solving  examples  in  simple 
proportion  without  making  the  statement  in  form^  may  be  used 
by  those  who  prefer  it. 

Every  question  in  simple  proportion  gives  tlirce  terms  to  find  a 
fourth.  Of  the  three  given  terms,  two  will  always  be  like  numbers, 
forming  the  complete  ratio,  and  the  third  will  be  of  the  same  name  or 
kind  as  the  required  term,  and  may  be  regarded  as  the  antecedent  of 
the  incomplete  ratio ;  hence  the  required  term  may  be  found  by  mul- 
tiplving  this  third  term,  or  antecedent,  by  the  ratio  of  the  other  two, 
(421,11). 

From  the  conditions  of  the  question  we  can.  readily  determine 
whether  the  answer,  or  required  term,  will  be  greater  or  less  than  the 
third  term ;  if  greater,  then  the  ratio  will  be  greater  than  1,  and  the 
two  like  numbers  must  be  arranged  in  the  form  of  an  improper  frac- 
tion, as  a  multiplier ;  if  less,  then  the  ratio  will  be  lesS  than  1,  and 
the  two  like  numbers  must  be  arranged  in  the  form  of  a  proper  frac- 
tion, as  a  multiplier. 

1.  If  4  tons  of  hay  cost  $3G^  what  will  5  tons  cost? 

OPERATION.  Analysis.      In  this  example,  4 

$36  X  -  =  §45    Ans.  "tons  and  5  tons  are  the  like  terms, 

and  $36  is  the  third  term,  and  of 
the  same  kind  as  the  answer  sought.  Now  if  4  tons  cost  $36,  will  5 
tons  cost  more,  or  less,  than  $36  ?  Evidently  more  :  and  the  required 
term  will  be  greater  than  the  third  term,  $36,  and  the  ratio  greater 
than  1.  AVe  therefore  arrange  the  like  terms  in  the  form  of  an  im- 
proper fraction,  |,  for  a  multiplier,  and  obtain  $45,  the  answer. 

2.  If  7  men  build  21  rods  of  wall  in  a  day,  how  many  rods  will 
4  men  build  in  the  same  time  ? 

OPERATION.  Analysis.     In  this  example,  7 

2X  X  -  =  12  rods    Ans.  ^^^  ^^^  ^  ^^^^^  ^^^  ^^^^  ^^^^  terms, 

and  21  rods  is  the  third  term,  and 
of  the  same  kind  as  the  answer  sought.  Since  4  men  will  perform  less 
work  than  7  men  in  the  same  time,  the  required  term  will  be  less  than 


252  PROPORTION. 

21,  and  the  ratio  less  than  1.     We  therefore  arrange  the  like  terms 

in  the  form  of  a  proper  fraction,  4,  and  obtain  by  multiplication,  12 

rods,  the  answer. 

4:36.    Hence  the  following 

Rule.     I.  With  the  two  given  numbers ^  icMch  are  of  the  same 

name  or  Mnd^  form  a  ratio  greater  or  less  than  1,  according  as  the 

answer  is  to  he  greater  or  less  than  the  third  given  number. 

{      II.   Multiple/  the  third  number  by  this  ratio  ;  the  product  will  be 

,  the  required  number  or  answer. 

Notes. — 1.  Mixed  numbers  should  first  be  reduced  to  improper  fractions,  and 
the  ratio  of  the  fractions  found  according  to  418. 

2.  Reductions  and  cancellation  may  be  applied  as  in  the  first  method. 

The  following  examples  may  be  solved  by  either  of  the  fore- 
going methods. 

EXAMPLES   rOR   PRACTICE. 

1.  If  12  gallons  of  wine  cost  $80,  what  will  68  gallons  cost?  /e)  / 

2.  If  9  bushels  of  wheat  make  2  barrels  of  flour,  how  many 
barrels  of  flour  will  100  bushels  make  ?  Aiis.  22|. 

8.  If  18  bushels  of  wheat  be  bought  for  $22.25,  and  sold  for 
$26.75,  how  much  will  be  gained  on  240  bushels,  at  the  same  rate 
of  profit  ?  Ans.  $60. 

4.  If  6 J  bushels  of  oats  cost  $8,  what  will  9i  bushels  cost?  'i-^ 

5.  What  will  87.5  yards  of  cloth  cost,  if  1|  yards  cost  $.42  ?  9.), 

6.  If  by  selling  $1500  worth  of  dry  goods  I  gain  $275.40,  what 


amount  must  I  sell  to  gain  $1000  ?^// 

7.  If  20  men  can  perform  a  piece  of  work  in  15  days,  how 
many  men  must  be  added  to  the  number,  that  the  work  may  be 
accomplished  in  |  of  the  time  ?  Ans.  5. 

8.  If  100  yd.  of  broadcloth  cost  $473.07^3,  how  much  will 
3.25  yd.  cost?  f  /<:^7  (  V  O'O  y^  fV 

9.  If  1  lb.  4  oz.  10  pwt.  of  gold  may  be  bought  for  $260.70, 
how  much  may  be  bought  for  $39.50  ?  Ans.  2  oz.  10  pwt. 

10.  In  what  time  can  a  man  pump  54  barrels  of  water,  if  he 
pump  24  barrels  in  1  h.  14  min.  ?        Ans.  2  h.  46  min.  30  sec. 

11.  If  I  of  a  bushel  of  peaches  cost  $^|,  what  part  of  a  bushel 
can  be  bought  for  $3/^  ?  "  A  ns.  ^^  bu.' 


COMPOUND  PROPORTION.  253 

12.  If  the  annual  rent  of  46  A.  3  E,  14  P.  of  land  be  $374.70, 
how  much  will  be  the  rent  of  35  A.  2  E.  10  P.  ?  ^  - 

13.  If  a  man  gain  §1870.65  by  his  business  in  1  yr.  3  mo.,  how 
much  would  he  gain  in  2  yr.  8  mo.,  at  the  same  rate  ?J  y  f(^ w^j 

14.  Two  numbers  are  to  each  other  as  5  to  7},  and  the  less  is 
164.5,  what  is  the  greater?  Ans.  246.75. 

15.  If  16  head  of  cattle  require  12  A.  3  E.  36  P.  of  pasture 
during  the  season,  how  many  acres  will  132  head  of  cattle  require  ? 

Ans.  107  A.  7  P. 

16.  If  a  speculator  in  grain  gain  $26.32  by  investing  $325,  how 
much  would  he  gain  by  investing  $2275?/  ^"^^  ,,  . 

17.  What  will  be  the  cost  of  paving  an  open  court  60.5  ft.  long 
and  44  ft.  wide,  if  14.25  sq.  yd.  cost  $34^?  ;_ 

18.  At  6i  cents  per  dozen,  w^hat  will  be  the  cost  of  10  J  gross 
of  steel  pens?    V^^  /^  ,      S-fh 

19.  If  when  wheat  is  7s.  6d.  per  bushel,  the  bakers'  loaf  will 
weigh  9  oz.,  what  ought  it  to  weigh  when  wheat  is  6s.  per  bushel? 

Ans.  Hi  oz. 


I 


COMPOUND  PROPORTION. 

437.  Compound  Proportion  is  an  expression  of  equality  be- 
tween a  compound  and  a  simple  ratio,  or  between  two  compound 
ratios. 

It  embraces  the  class  of  questions  in  which  the  causes,  or  the 
effects,  or  both,  are  compound.  The  required  term  must^e  either 
a  simple  cause  or  effect,  or  a  single  element  of  a  compound  cause 
or  effect. 

FIRST   METHOD. 

1.  If  8  men  mow  40  acres  of  grass  in  3  days,  how  many  acres 
will  9  men  mow  in  4  days  ? 

STATEMENT. 
1st  cause.        2d  cause.  1st  efifect.         2d  effect. 


(3=        14      =      40 

:      (?) 

Or,  8  X  3  :  9  X  4      =      40 

■      (?) 

22 

254 


PROPORTION. 


OPERATION. 


(n-m^.m...... 


Analysis.      In  this  ex- 
ample the    required   term 
is  the  second  effect ;  and  the 
statement  is,  8  men  3  days 
is  to  9  men  4  days,  as  40  acres  is  to  (  ?  ),  or  how  many  acres.    Dividing 
the  continued  product  of  all  the  elements  of  the  means  by  the  ele- 
ments of  the  given  extreme,  we  obtain  (  ?  )  =r  60  acres. 

2.  If  6  compositors  in  14  hours  can  set  36  pages  of  56  lines 
each,  how  many  compositors^  in  12  hours^  can  set  48  pages  of  54 
lines  each  ? 

STATEMENT. 
1st  cause.  2d  cause.        1st  effect.        2d  effect. 


f    6    .     I    (?)  ..f    36    .     I 
1 14    •     1     12    ••!    56    •     X 


48 
54 


OPERATION. 

(?) 


Analysis.  In  this  example,  an  element  of 
the  second  cause  is  required ;  and  the  state- 
ment is,  6  compositors  14  hours  is  to  (  ?)  com- 
positors 12  hours  as  36  pages  of  56  lines  each 
is  to  48  pages  of  54  lines  each.  Now,  since  the 
(  )  =  9  Ans.  required  term  is  an  element  of  one  of  the  means, 
we  divide  the  continued  product  of  all  the  ele- 
ments of  the  extremes  by  the  continued  product  of  all  the  given  ele- 
ments of  the  means.  Placing  the  dividend  on  the  right  of  the  verti- 
cal line  and  the  divisors  on  the  left,  and  canceling  equal  factors  we 
obtain  (  ?  )  ==  9. 

4:3§.   From  these  illustrations  we  deduce  the  following 

Hule.^    I.    Of  the  given  tcrmSy  select  those  which  constitute  the 

causes,  and  those  which  constitute  the  effects,  and  arrange  them  in 

couplets, putting  (?  ^  in  place  of  the  required  term. 

II.    Then,  if  the  hlank  term  (?)  occur  in  either  of  the  extremeSj 

divide  the  j)'^oduct  of  the  means  hy  the  product  of  the  extremes; 

hut  if  the  hlanh  term  occur  in  either  mean,  divide  the  product  of 

the  extremes  hy  the  product  of  the  means. 

I^OTKS.  —  1.  The  causes  must  "he  exactly  alike  in  the  »innj6erand  hhxd  of  their 
terms  :   the  same  is  true  of  the  effects. 

2.  The  same  preparation  of  the  terms  by  reduction  is  to  be  observed  as  in 
simple  proportion. 


f 


■I    the 

L 


COMPOUND  PllOPOETIOK.  255 

SECOND    METHOD. 

4l«l^.  The  second  method  given  in  Simple  Proportion^  is  also 
applicable  in  Compound  Proportion. 

In  every  example  in  compound  proportion  all  the  terms  appear  in 
couplets,  except  one,  called  the  odd  term,  which  is  always  of  the  same 
kind  as  the  answer  sought.  Hence  the  required  term  in  a  compound 
proportion  may  be  found,  by  multiplying  the  odd  term  by  the  com- 
pound ratio  composed  of  all  the  simple  ratios  formed  by  these  couplets, 
each  couplet  being  arranged  in  the  form  of  a  fraction. 

The  fraction  formed  by  any  couplet  will  be  improper  when  the  re- 
quired term,  considered  as  depending  on  this  couplet  alone,  should 
be  greater  than  the  odd  term ;  and  proper,  when  the  required  term 
should  be  less  than  the  odd  term. 

1.  If  it  cost  84320  to  supply  a  garrison  of  32  men  with  pro- 
visions for  18  days,  when  the  rations  are  15  ounces  per  day,  what 
will  it  cost  to  supply  a  garrison  of  24  men  34  days,  when  the 
rations  are  12  ounces  per  day  ? 

OPERATION, 
men.  days.        ounces. 

§4320   X   M   X   II   X   If  =  $4896 

BY  CANCELLATION.  ANALYSIS.     In   this  example  there  are 

4320  three  pairs  of  terms,  or  couplets,  viz.,  32 

24  men  and  24  men,  18  days  and  34  days,  15 

34  ounces  and  12  ounces ;  and  there  is  an  odd 

12  term,  $4320,  which  is  of  the  same  kind  as 

(  )  =  $4896  Arts,  the  required  term.  We  arrange  each  coup- 
let as  a  multiplier  of  this  term,  thus; 
First,  if  it  cost  $4320  to  supply  32  men,  will  it  cost  more,  or  less,  to 
supply  24  men  ?  Less  ;  we  therefore  arrange  the  couplet  in  the  form  of 
a  proper  fraction  as  a  multiplier,  and  we  have  $4320  X  ^4.  Next,  if  it 
cost  $4320  to  supply  a  garrison  18  days,  will  it  cost  more,  or  less,  to 
supply  it  34  days?  More ;  hence  the  multiplier  is  the  improper  frac- 
tion ^5,  and  we  have  $4320  X  Jf  X  f^  Next,  if  it  cost  $4320  to 
supply  a  garrison  with  rations  of  15  ounces,  will  it  cost  more,  or  less, 
Avhen  the  rations  are  12  ounces?  Less;  consequently,  the  multiplier 
is  the  proper  fraction  \Z,  and  we  have  $4320  X  Jl  X  ?^  X  j^  =$4896, 
the  required  term.     Hence  the  following 


(0 

32 

18 

15 

256  PROPORTION. 

Rule.  I.  Of  the  terms  com2?osing  each  couplet  form  a  ratio 
greater  or  less  than  1,  in  the  same  manner  as  if  the  aiisicer  de- 
pended on  those  two  and  the  third  or  odd  term. 

II.  Multiply  together  the  third  or  odd  tei^i  and  these  ralios ; 
the  product  will  he  the  ansiver  sought. 

EXAMPLES    FOR   PRACTICE. 

1.  If  12  horses  plow  11  acres  in  5  days,  how  many  horses 
would  plow  33  acres  in  18  days?  Ans.  10. 

2.  If  480  bushels  of  oats  will  last  24  horses  40  days,  how  long 
will  300  bushels  last  48  horses,^  at  the  same  rate  ? 

Ans.  12-2  days. 

3.  If  7  reaping  machines  can  cut  1260  acres  in  12  days,  in 
how  many  days  can  16  machines  reap  4728  acres  ? 

Ans.  19.7  days. 

4.  If  144  men  in  6  days  of  12  hours  each,  build  a  wall  200  ft. 
long,  3  ft.  high,  and  2  ft.  thick,  in  how  many  days  of  7  hours 
each  can  30  men  build  a  wall  350  ft.  long,  6  ft.  high,  and  3  ft. 
thick?  Ans.   259.2  da. 

5.  In  how  many  days  will  6  persons  consume  5  bu.  of  potatoes, 
if  3  bu.  3  pk.  last  9  persons  22  days  ? 

6.  How  many  planks  lOf  ft.  long  and  IJ  in.  thick,  are  equiva- 
lent to  3000  planks  12  ft.  8  in.  long  and  2$  in.  thick? 

Aus.  6531}. 

7.  If  300  bushels  of  wheat  @  §1.25  will  discharge  a  certain 
debt,  how  many  bushels  @  §.90  will  discharge  a  debt  3 -times  as 
great?  Ans.  1256  bu. 

8.  If  468  bricks,  8  inches  long  and  4  inches  wide,  are  required 
for  a  walk  26  ft.  long  and  4  ft.  wide,  how  many  bricks  will  be 
required  for  a  walk  120  ft.  long  and  6  ft.  wide  ? 

9.  If  a  cistern  17J  ft.  long,  lOJ  ft.  wide,  and  13  ft.  deep,  hold 
546  barrels,  how  many  barrels  will  a  cistern  hold  that  is  16  ft. 
long,  7  ft.  wide,  and  15  ft.  deep?  Ans.  384  bbK 

10.  If  11  men  can  cut  147  cords  of  wood  in  7  days,  when  they 
work  14  hours  per  day,  how  many  days  will  it  take  5  men  to  cut 
150  cords,  working  10  hours  each  day? 


I 


PROMISCUOUS  EXAMPLES.  257 

PROMISCUOUS   EXAMPLES   IN   PROPORTION. 

1.  If  a  staff  4  ft.  long  cast  a  shadow  7  ft.  in  length,  what  is 
the  hight  of  a  tower  that  casts  a  shadow  of  198  ft.  at  the  same 
time?  Ans.  113^  ft. 

2.  A  person  failing  in  business  owes  $972,  and  his  entire  prop- 
erty is 'Worth  but  $607.50;  how  much  will  a  creditor  receive  on  a 
debt  of  $11.33J?  Ans.  $7.08+. 

3.  If  3  cwt.  can  be  carried  660  mi.  for  $4,  how  many  cwt.  can 
be  carried  60  mi.  fur  $12  ?  Ans.  99. 

4.  A  man  can  perform  a  certain  piece  of  work  in  18  days  by 
working  8  hours  a  day;  in  how  many  days  can  he  do  the  same 
work  by  working  10  hours  a  day?  Arts.  14|. 

5.  How  much  land  worth  $16.50  an  acre,  should  be  giv  n  in 
exchange  for  140  acres,  worth  $24.75  an  acre? 

6.  If  I  gain  $155.52  on  $1728  in  1  yr.  6  mo.,  how  much  will 
I  gain  on  $750  in  4  yr.  6  mo.?  Ans.  $202.50. 

7.  If  1  lb.  12  oz.  of  wool  make  2  J  yd.  of  cloth  6  qr.  wide,  how 
many  lb.  of  wool  will  it  take  for  150  yd.  of  cloth  4  qr.  wide  ? 

8.  What  number  of  men  must  be  employed  to  finish  a  piece  of 
work  in  5  days,  which  15  men  could  do  in  20  days  ?     Ans.  60. 

9.  At  12s.  7d.  per  oz.,  N.  Y.  currency,  what  will  be  the  cost 
of  a  service  of  silver  plate  weighing  15  lb.  11  oz.  13  pwt.  17  gr.  ? 

10.  If  a  cistern  16  ft.  long,  7  ft.  wide,  and  15  ft.  deep,  cost 
$36.72,  how  much,  at  the  same  rate  per  cubic  foot,  would  another 
cistern  cost  that  is  17i  ft.  long,  10}  ft.  wide,  and  16  ft.  deep? 

11.  A  borrows  $1200  and  keeps  it  2  yr.  5  mo.  5  da.;  what 
sum  should  he  lend  for  1  yr.  8  mo.  to  balance  the  favor  ? 

12.  A  farmer  has  hay  worth  $9  a  ton,  and  a  merchant  has  flour 
worth  $5  per  barrel.  If  in  trading  the  former  asks  $10.50  for 
his  hay,  how  much  should  the  merchant  ask  for  his  flour  ? 

13.  If  12  men,  working  9  hours  a  day  for  15|  days,  were  able 
to  execute  §  of  a  job,  how  many  men  may  be  withdrawn  and  the 
job  be  finished  in  15  days  more,  if  the  laborers  are  employed 
only  7  hours  a  day  ?  Ai2s.  4. 

14.  If  the  use  of  $300  for  1  yr.  8  mo.  is  worth  $30,  how  much 
is  the  use  of  $210.25  for  3  yr.  4  mo.  24  da.  worth? 

22*  B 


258  PROPORTION. 

15.  What  quantity  of  lining  f  yd.  wide,  will  it  require  to  line 
91  yd.  of  cloth,  1}  yd.  wide  ?  ,        Ans.  15|  yd. 

IG.  If  it  cost  $95.60  to  carpet  a  room  24  ft.  by  18  ft.,  how 
much  will  it  cost  to  carpet  a  room  38  ft.  by  22  ft.  with  the  same 
material?  Ans.  $185.00+. 

17.  If  IGj'^g  cords  of  wood  last  as  long  as  Il^^g  tons  of  coal, 
how  many  cords  of  wood  will  last  as  long  as  I5/3  tons  of  coal? 

18.  A  miller  has  a  bin  8  ft.  long,  44  ft.  wide,  and  2i  ft.  deep, 
and  its  capacity  is  75  bu. ;  how  deep  must  he  make  another  bin 
which  is  to  be  18  ft.  long  and  3|  feet  wide,  that  its  capacity  may 
be  450  bu.  ?  Ans.  7/3  ft. 

19.  If  4  men  in  2  J  da3^s,  mow  6 J  acres  of  grass,  by  working 
81  hours  a  day,  how  many  acres  will  15  men  mow  in  3|  days,  by 
working  9  hours  a  day?  Ans.  40}^  acres. 

20.  If  an  army  of  600  men  have  provisions  for  5  weeks,  allowing 
each  man  12  oz.  a  day,  how  many  men  may  be  maintained  10 
weeks  with  the  same  provisions,  allowing  each  man  8  oz.  a  day  ? 

21.  A  cistern  holding  20  barrels  has  two  pipes,  by  one  of  which 
it  receives  120  gallons  in  an  hour,  and  by  the  other  discharges 
80  gallons  in  th%  same  time ;  in  how  many  hours  will  it  be  filled  ? 

22.  A  merchant  in  selling  groceries  sells  \^j%  oz.  for  a  pound; 
how  much  does  he  cheat  a  customer  who  buys' of  him  to  the  amount 
of  $38.40  ?  Ans.  $3.45. 

23.  If  5  lb.  of  sugar  costs  $.62  J,  and  8  lb.  of  sugar  are  worth 
5  lb.  of  coffee,  how  much  will  75  lb.  of  cofiee  cost? 

24.  B  and  C  have  each  a  farm;  B's  farm  is  worth  $32.50  an 
acre,  and  C's  $28.75 ;  but  in  trading  B  values  his  at  $40  an  acre. 
What  value  should  C  put  upon  his  ? 

25.  If  it  require  859|  reams  of  paper  to  print  12000  copies  of 
an  8vo.  book  containing  550  pages,  how  many  reams  will  be  required 
to  print  3000  copies  of  a  12mo.  book  containing  320  pages? 

26.  If  248  men,  in  5  J  days  of  12  hours  each,  dig  a  ditch  of  7 
degrees  of  hardness,  232i  yd.  long,  3f  yd.  wide,  and  2  J  yd.  deep; 
in  how  many  days  of  9  hours  each,  will  24  men  dig  a  ditch  of  4 
degrees  of  hardness,  387i  yd.  long,  5^  yd.  wide,  and  3  J  yd  deep? 

Ans.  155 


NOTATION. 


259 


PERCENTAGE. 

4:4:0.  Per  Cent,  is  a  contraction  of  the  Latin  phrase  per 
centum,  and  signifies  hy  the  hundred ;  that  is,  a  certain  part  of 
every  hundred,  of  any  denomination  whatever.  Thus,  4  per  cent 
means  4  of  every  hundred,  and  may  signify  4  cents  of  every  100 
cents,  4  dollars  of  every  100  dollars,  4  pounds  of  every  100 
pounds,  etc. 

NOTATION. 

44^.  The  character,  %,  is  generally  employed  in  business 
transactions  to  represent  the  words  per  cent. ;  thus  G  %  signifies 
G  per  cent. 

44^.  Since  any  per  cent,  is  some  number  of  hundredths,  it  is 
properly  expressed  by  a  decimal  fraction;  thus  5  per  cent. 
=z  b  (fo  =■  -05.  Per  cent,  may  always  be  expressed,  however, 
either  by  a  decimal  or  a  common  fraction ,  as  shown  in  the  following 


TABLE. 


Words. 

1  per  cent.  = 

2  per  cent.  = 
4  per  cent.  = 
6    per  cent.  =» 

6  per  cent.  = 

7  per  cent.  = 

8  per  cent.  = 
10  per  cent.  = 
20  per  cent.  = 
25  per  cent.  = 
50    per  cent.  = 

100    per  cent.  = 

125    per  cent.  = 

i  per  cent.  = 

£  per  cent.  = 

12^  per  cent.  = 


Syinlx)ls. 

1  % 

2  % 

4  /. 

5  % 

6  % 

7  % 

8  % 

10  % 

20  % 

25  % 

50  % 

100  % 

125  % 


1  t/     — 


12^  fc 


Decimals. 

.01 

.02 

.04 

.05 

.06 

.07 

.08 

.10 

.20 

.25 

.50 
1.00 
1.25 

.001 

.OOJ 

.121 


Common  trac 

tions. 

ih 

= 

Toff 

T?ir 

= 

oV 

I'U 

= 

o-S 

5 
lOJS 

= 

A 

6 
JUS 

== 

i' 

7 

= 

Tffff 

• 
TUff 

» 

/j 

1  0 

ro(5 

= 

iV 

2f> 
TOCT 

= 

i 

?5 

= 

i 

5ft 

TiJff 

== 

i 

1  00 

150 

= 

I 

Tft(r 

foTJ 

= 

i 
f 

1 

25ff 

T6^ 

= 

i 

260  PERCENTAGE, 

EXAMPLES   FOR   PRACTICE. 

1.  Express  de 
16 


EXAMPLES    FOR   PRACTICE. 

1.  Express  decimally  3  per  cent. ;  9  per  cent. ;  12  per  cent. ; 
xu  per  cent.;  23  per  cent.;  37  per  cent.;  75  per  cent.;  125  per 
cent. ;  184  per  cent. ;  205  per  cent. 

2.  Express  decimally  15  %  ;  H  %  ;  4i  %  ;  5i  %  ;  8J  %  ; 
20i  %  ;  251  %  ;  35|  %  ;  24f  %  ;  130J  %. 

3.  Express  decimally  i  per  cent. ;  f  per  cent. ;  i  per  cent. ; 
I  per  cent. ;  |  per  cent. ;  /^  per  cent. ;  j3_6^  per  cent. ;  If^  per 
cent. ;  10-1-  per  cent. 

4.  Express  by  common  fractions,  in  their  lowest  terms,  4  % ; 
87J  %  ;  16|  %  ;  ll^  %  ;  42|  % ;  45/^  %  ;  43/,  %. 

5.  What  per  cent,  is  .0725  ? 

Analysis.    .0725  =  .071  =  7i  ^,  Ans. 

6.  What  per  cent,  is  .065?  Ans.  6  J  %. 

7.  What  per  cent,  is  .14375?  Ans,  14i  %. 

8.  What  per  cent,  is  .0975  ? 

9.  What  per  cent,  is  .014  ? 

10.  What  per  cent,  is  .1025  ? 

11.  What  per  cent,  is  .004? 

12.  What  per  cent,  is  .028  ? 

13.  What  %  is  .1324? 

14.  What  %  is  .084f  ? 

15.  What  %  is  .004-j\?  Ans.  /-j-  % 

16.  What  %  is  .003  J^  ? 


GENERAL  PROBLEMS  IN  PERCENTAGE. 

4:43.  In  the  operations  of  Percentage  there  are  five  parts  or 
elements,  namely  :  Rate  per  cent.,  Percentage,  Base,  Amount,  and 
Difference. 

4:44«  Hate  per  Cent.,  or  Rate,  is  the  decimal  which  denotes 
how  many  hundredths  of  a  number  are  to  be  taken. 

Notes. — 1.  Such  expressions  as  6  per  cent.,  and  5  "^.are  essentinlly  dechnah, 
the  words  ^icr  cent.,  or  the  character  % ,  indicating  the  decimal  denominator. 

2.  If  the  decimal  bo  reduced  to  a  common  fraction  in  its  loioent  terms,  this 
fraction  will  still  be  the  equivalent  ratef  though  not  the  rate^«r  cent. 


PROBLEMS  IN  PERCENTAGE. 


261 


I 


44d,   Percentage  is  that  part  of  any  number  wliicli  is  indi- 
cated by  the  rate. 

446.  The  Base  is  the  number  on  which  the  percentage  is 
computed. 

447.  The  Amount  is  the  sum  obtained  by  adding  the  per- 
centage to  the  base. 

448.  The  Difference  is  the  remainder  obtained  by  subtract- 
ing the  percentage  from  the  base. 

PROBLEM   I. 

449.  Given,  the   base   and  rate,  to   find  the   per- 
ntage. 
1.  What  is  5  %  of  360  ? 


centage, 


Rule. 

Note  1.- 
f actor  8. 


Analysis.  Since  5  ^  of  any 
number  is  .05  of  that  number, 
(442),  we  multiply  the  base,  360, 
by  the  rate,  .05,  and  obtain  the 
percentage,  18.  Or,  since  the  rate 
is  j^iy  =  oV*  ^^^  h.^^^  3G0  X  2V  == 
18,  the  percentage.  Hence  the  fol- 
lowing 

Multiply  the  hase  hy  the  rate. 
Percentage  is  always  a  product,  of  which  the  base  and  rate  ar©  the 


OPERATION, 

860 
__.05 

18.00,  Ans, 

Or, 

860  X  2^5  =  18,  Ans, 


1.  What 

2  What 

3.  What 

4.  What 

5.  What 

6.  What 

7.  What 

8.  What 

9.  What 
10.  What 


EXAMPLES    FOR   PRACTICE. 

s  4  per  cent,  of  250  ? 
s  7  per  cent,  of  3500  ? 
s  16  per  cent,  of  324  ? 
s  12^  per  cent,  of  $5600  ? 
s  9  %  of  785  lbs.  ? 
s  25  %  of  960  mi.? 
s  75  %  of  487  bu.  ? 
s38i  %  of  2757  men? 

125  %  of  756  ? 

I  %  of  $2864  ? 


Ans.  10. 

Ans.  245. 

Ans.  51.84. 

A71S.  §700. 


Ans.  865.25  bu. 


Ans.  $5.91. 


_1 
1  tiU* 


262  PERCENTAGE. 

11.  What  is  3|  %  of  $856?  Ans.  §31.39 

12.  What  is  I  %  of  I  ?  Ans. 

13.  What  is  14f   %  of  di  ? 

1-1.  If  the  base  is  §375,  and  the  rate  .05,  what  is  the  percent- 
age ?  Ans.  $18.75. 

15.  A  man  owed  §536  to  A,  $450  to  B,  and  $784  to  C;  how 
much  money  will  be  required  to  pay  54  %  of  his  debts  ? 

16.  My  salary  is  $1500  a  year;  if  I  pay  15  %  for  board,  5  % 

for  clothing,  6  %  for  books,  and  8  %   for  incidentals,  what  are 

my  yearly  expenses  ?  Ans.  $510. 

NoTK  2.  — 15  ^  +  5  %  +  (>  %  +8  %  =Zi  %.  In  all  cases  where  several 
rates  refer  to  the  same  base,  they  may  be  added  or  subtracted,  according  to  the 
conditions  of  the  question. 

17.  A  man  having  a  yearly  income  of  $3500,  spends  10  per 
cent  of  it  the  first  year,  12  per  cent,  the  second  year,  and  18  per 
cent,  the  third  year;  how  much  does  he  save  in  the  3  years? 

18.  A  had  $6000  in  a  bank.  He  drew  out  25  %  of  it,  then 
SO  %  of  the  remainder,  and  afterward  deposited  10  %  of  what 
he  had  drawn ;  how  much  had  he  then  in  bank  ?    Ans.  $3435. 

19.  A  merchant  commenced  business,  Jan.  1,  with  a  capital  of 
$5400,  and  at  the  end  of  1  year  his  ledger  showed  the  condition 
of  his  business  as  follows :  For  Jan.,  2  %  gain  ;  Feb.,  3}  %  gain  ; 
March,  }  %  loss;  Apr.,  2  %  gain;  May,  2}  %  gain;  June,  If 
%  loss;  July,  1}  %  gain;  Aug.,  1  %  loss;  Sept.,  2|  %  gain; 
Oct.,  4  %  gain;  Nov.,  J  %  loss;  Dec,  3  ^  gain.  What  were 
the  net  profits  of  his  business  for  the  year?  Ans.  §918. 

ruoBLE:\i  II., 
450.   Given,  the   percentage    and   baso^  to  find   the 
rate, 

1.  What  per  cent  of  360  is  18  ? 

opERATiox.  Analysis.    Since  the  pcrcent- 

18  -i-  360  =  .05  =  5  %  age  is  always  the  product  of  the 


Qj.  base  and  rate,  (449),  we  divide 

=  .05  =  5  %  *'       * 

required  ratQ,  .05  =?  5  fo.     Hence  the 


,8   1    f).^ P.  c/  the  given  percentage,  18,  b}^  the 

■3  0^  —  -iU  —  •       —       /o  g.^^j^  i^^g^^  3QQ^  and.  obtain  the 


PROBLEMS  IN  PERCENTAGE.  263 

Rule.     Divide  the  percentage  hy  the  base. 

EXAMPLES    FOR   PRACTICE. 

1.  What  per  cent,  of  $720  is  $21.60  ?  Ans.  3. 

2.  What  per  cent,  of  1500  lb.  is  234  lb.  ? 

3.  What  per  cent,  of  980  rd.  is  40  rd.  ? 

4.  What  per  cent,  of  £320  10s.  is  £25  12.8s.  ?         Ans.  8. 

5.  What  per  cent,  of  46  gal.  is  5  gal.  3  qt.?  Ans.   12  J. 

6.  What  per  cent,  of  7.85  mi.  is  5.495  mi.?       *    Ans.  70. 

7.  What  per  cent,  of  j%  is  |  ?  Ans.  75. 

8.  What  per  cent,  of  ^  is  t>^^  ? 

9.  What  per  cent,  of  560  is  80  ? 

10.  The  base  is  $578,  and  the  percentage  is  $26.01 ;  what  is  the 
rate?  Ans.  4i  %. 

11.  The  base  is  $972.24,  and  the  percentage  is  §145.836;  what 
is  the  rate  ? 

12.  An  editor  having  5600  subscribers,  lost  448;  what  was  his 
loss  per  cent?  Ans.   8. 

13.  A  merchant  owes  $7560,  and  his  assets  are  $4914;  what 
per  cent,  of  his  debts  can  he  pay  ?  Ans,   65. 

14.  A  man  shipped  2600  bushels  of  grain  from  Chicago,  and 
455  bushels  were  thrown  overboard  during  a  gale ;  what  was  the 
rate  per  cent,  of  his  loss  ? 

15.  A  miller  having  720  barrels  of  flour,  sold  288  barrels;  what 
per  cent,  of  his  stock  remained  unsold  ?  Ans.  60. 

16.  AVhat  per  cent,  of  a  number  is  30  %  of  |  of  it  ? 

17.  The  total  expenditures,  of  the  General  Government,  for  the 
year  ending  June  30,  1858,  were  $83,751,511.57;  the  expenses 
of  the  War  Department  were  $23,243,822.38,  and  of  the  Navy 
Department,  $14,7 12, 610.21.     What  per  cent,  of  the  whole  cx- 

^jense  of  government  went  for  armed  protection  ? 

^^B  Ans.  45J-,  nearly. 

18.  In  the  examination  of  a  class,  165  questions  were  sub- 
mitted to  each  of  the  5  members;  A  answered  130  of  them,  B 
125,  C  96,  D  110,  and  E  160.  What  was  the  standing  of  the 
class?  Ans.  75.27  %. 


264  PERCENTAGE. 


PROBLEM  III. 

431.    Given,  the   percentage   and  rate,  to  find  the 
base. 

1.   18  is  5  %  of  what  number? 

.^x,,T,.mr^^r  Analysis.     SmcG   tliG  pGrceiit- 

OPERATIOX.  .  ^ 

age  is  always  the  product  of  the 
18  -^  .Oo  =  360,  Arts.  ^^^^  ^^^  ^^^^^  ^^^^^^  ^^  ^j^j^^ 

Or,  the  given  percentage,  18,  by  the 

18  -7-  3j^0  =  360,  Ans.  given  rate,  .05,  or  J^,  and  obtain 

the  base,  360.     Hence  the 
Rule.     Divide  the  percentage  hy  the  rate. 


EXAMPLES   FOR   PRACTICE. 

1.  ]  8  is  25  %  of  what  number  ?  Ans.  72. 

2.  54  is  15  %  of  what  number? 

3.  17.5  is  2J  cjo  of  what  number?  Ans.  750. 

4.  2.28  is  5  %  of  what  number? 

5.  414  is  120  %  of  what  number  ? 

6    6119  is  105}  cjo  of  what  number?  Ans,   5800. 

7.  .43  is  71f  ^0  of  what  number?  Am.  .6. 

8.  The  percentage  is  $18.75,  and  the  rate  is  2}  %  ;  what  is  the 
base  ?  Ans,  $750. 

9.  The  percentage  is  31},  and  the  rate  31}  %  ;  what  is  the 
base  ? 

10.  I  sold  my  house  for  $4578,  which  was  84  %  of  its  cost; 
what  was  the  cost  ?  Ans,  $5450. 

11.  A  wool  grower  sold  3150  head  of  sheep,  and  had  30  %  of 
his  original  flock  left )  how  many  sheep  had  he  at  first  ? 

12.  A  man  drew  40  %  of  his  bank  deposits,  and  expended  13  J 
%  of  the  money  thus  drawn  in  the  purchase  of  a  carriage  worth 
$116;  how  much  money  had  he  in  bank?  Ans.  $2175. 

13.  If  $147.56  is  13|  %  of  A's  money,  and  4f  %  of  A's 
money  is  8  %  of  B's,  how  much  more  money  has  A  than  B  ? 

Ans.  $461.12}. 


I 


PROBLEMS  IN  PERCENTAGE.  265 

4.  In  a  battle  4  %  of  the  army  were  slain  upon  the  field  :|itid  5 

%  of  the  remainder  died  of  wounds,  in  the  hospital.     The  difier- 

ence  between  the  killed  and  the  mortally  wounded  was  168 ;  how 

many  men  were  there  in  the  army  ?  Ans.  21000. 

Note.— 100  %  —  4  ^  =  96  %,  left  after  the  battle;  and  5  %  of  96  "2^  =- 
4 J  %,  the  part  of  the  army  that  died  of  wounds. 

15.  A  owns  f  of  a  prize  and  B  the  remainder;  after  A  has 
taken  40  %  of  his  share,  and  B  20  %  of  his  share,  the  remainder 
is  equitably  divided  between  them  by  giving  A  $1950  more  than 
B ;  what  is  the  value  of  the  prize  ?  Ans.  $7800. 

PROBLEM   IV.  . 

433.   Given,  the  amount  and  rate,  to  find  tlie  base. 
1.  What  number  increased  by  5  %  of  itself  is  equal  to  378  ? 
OPERATION.  Analysis.      If    any   number 

1  4-      05  =  1  05  ^®  increased  by  5  ^  of  itself 

378  _^  1.05  =  360,  Ans,  *be  amount  will  be  1.05  times 

the  number.     We  therefore  di- 
Or,  vide  the  given  amount,  378,  by 

1  J.    1    *—     2  1  1.05,  or  |jl,  and  obtain  the  base. 

37g  -^  2  1  — _  350   ^^5.  360,   which  is  the  number  re- 

quired.    Hence  the 

BuLE.     Divide  the  amount  hy  1  jplns  the  rate. 

Note  1. — The  amount  is  always  a  product,  of  which  the  base  is  one  factor, 
and  1  plus  the  rate  the  other  factor. 

EXAMPLES    FOR   PRACTICE. 

1.  What  number  increased  by  15  %  of  itself  is  equal  to  644  ? 

Ans.  560. 

2.  A  has  ^815.36,  which  is  4  %  more  than  B  has;  how  much 
money  has  B  ?  Ans.  $784. 

3.  Having  increased  my  stock  in  trade  by  12  %  of  itself,  I 
find  that  I  have  $3800 ;  how  much  had  I  at  first  ? 

4.  In  1860  tiie  population  of  a  certain  city  was  39600,  which 
was  an  increase  of  10  %  during  the  10  years  preceding;  what 
was  the  population  in  1850  ? 

23 


2(56  PERCENTAGE. 

5.  Mj  crop  of  wheat  this  year  is  8  ^^  greater  than  my  crop  of 
last  year,  and  I  have  raised  during  the  two  years  5200  bushels; 
what- was  my  last  year's  crop?  Ans.  2500  bu. 

KoTE  2.  — 1.00+  1.08  =  2.08.     Hence,  5200  bu.  =  2.08  %  of  last  year's  crop. 

6.  The  net  profits  of  a  nursery  in  two  years  were  $6970,  and 
the  profits  the  second  year  were  5  %  greater  than  the  profits  the 
first  year ;  what  were  the  profits  each  year  ? 

Ans.  1st  year,  $3400 ;  2d  year,  $3570. 

7.  If  a  number  be  increased  8  %,  and  the  amount  be  increased 

7  foy  the  result  will  be  86.67 ;  required  the  number. 

KoTE  3. —  The  whole  amount  will  be  1.08  X  1.07  =  1.1556  times  the  original 
number. 

8.  A  produce  dealer  bought  grain  by  measure,  and  sold  it  by 
weight,  thereby  gaining  1}  %  in  the  number  of  bushels.  He 
sold  at  a  price  5  %  above  his  buying  price,  and  received  $4910.976 
for  the  grain  ;  required  the  cost.  Ans.  $4608. 

9.  B  has  6  %,  and  C  4  %  more  money  than  A,  and  they  all 
have  $11160  ;  how  much  money  has  A  ?  Ans.  $3600. 

'  10.  In  the  erection  of  a  house  I  paid  twice  as  much  for  mate- 
rial as  for  labor.  Had  I  paid  6  %  more  for  material,  and  9  %  more 
lor  labor,  my  house  would  have  cost  $1284 ;  what  was  its  cost  ? 

Ans.  $1200. 

PROBLEM   V. 

453.    Given,  tlie  difference  and  rate,  to  find  the  base. 

1.  What  number  diminished  by  5  %  of  itself,  is  equal  to  342  ? 

OPERATION.  Analysis.     If  any  number  be  di- 

-[ .05  =  .95  minished  by  5  %   of  itself,  the  dif- 

342  -f-  .95  =  360,  Ans.     fcrence  will  be  .95  of  the  number. 
Qr  We  therefore  divide  the  given  diiFer- 

1 ^1^   1=  19  ence,  342,  by  .95,  or  ^^,  and  obtain 

342  -I-  '  5  =  360,  Ans.     the  base,  3G0,  which  is  the,  requited 
number.     Hence  the  '; 

Hule.      Divide  the  difference  r^  1  minus  the  rate. 

XoTF. — The  difference  is  always  k  product,  of  which  the  base  is  one  factor, 
and  1  minus  the  rate  the  other. 


PROBLEMS  IN  PERCENTAGE.  2"f57 

EXAMPLES    FOR   PRACTICE. 

1.  "VYhat  number  diminished  by  10  %  of  itself  fs  equal  to  504? 

Atis.  5G0. 

2.  The  rate  is  8  ^,  and  the •  difference  $4.37;  what  is  the 
base? 

3.  After  taking  a^yay  15  %  of  a  heap  of  grain,  there  remained 
40  bu.  S^  pk. ;  how  many  bushels  were  there  at  first  ? 

An.9.  48  bu. 

4.  Having  sold  SG  %  of  my  land,  I  have  224  acres  left;  how 
much  land  had  I  at  first  ? 

5.  After  paying  65  %  of  my  debts,  I  find  that  62590  will  dis- 
charge the  remainder ;  how  much  did  I  owe  in  all  ? 

Ans.   $7400. 

6.  A  young  man  having  received  a  fortune,  deposited  80  ^^  of 
it  in  a  bank.  He  afterward  drew  20  %  of  his  deposit,  and  then 
had  §5760  in  bank;  what  was  his  entire  fortune? 

Ans.  ?9000. 

7.  A  man  owning  |  of  a  ship,  sold  12  %  of  his  share  to  A,  and 
the  remainder  to  B,  at  the  same  rate,  for  $20020;  what  was  the 
estimated  value  of  the  whole  ship  ?  Ans,  $26000. 

8.  An  army  which  has  been  twice  decimated  in  battle,  now 
contains  only  6480  men ;  what  was  the  original  number  in  the 
army?  Avs.  8000. 

^  9.  Each  of  two  men,  A  and  B,  desired  to  sell  his  horse  to  C. 
A  asked  a  certain  price,  and  B  asked  50  %  more.  A  then  re- 
duced his  price  20  %,  and  B  his  price  30  %,  at  which  prices  C 
took  both  horses,  paying  for  them  $148 ;  what  was  each  man's 
asking  B^?  4^^^     (A,     $^0. 

"^  ^      '     (B,  $120. 

^  "  10.  A  buyer  expended  equal  sums  of  money  in  the  purchase  of 
wheat,  corn,  and  oats.  In  the  sales,  he  cleared  6  %  on  the  wheat, 
and  3  %  on  the  corn,  but  lost  17  %  on  the  oats;  the  whole 
amount  received  was  $2336.  What  sum  did  he  lay  out  in  each 
kind  of  grain  ?  Ans,  $800. 


^68  PERCENTAGE. 

APPLICATIONS  OF  PERCENTAGE. 

4.54,  The  principal  applications  of  Percentage,  where  time  is 
not  considered,  are  Commission,  Stocks,  Profit  and  Loss,  Insurance, 
Taxes,  and  Duties.  And  since  the  five  problems  in  Percentage 
involve  all  the  essential  relations  of  the  parts  or  elements,  we  have 
for  the  above  applications  the  following 

General  Rule.  Note  what  elements  of  Percentage  are  given 
in  the  example^  and  what  elemerd  is  required ;  then  apply  the  spe- 
cial rule  for  the  corresponding  case. 

COMMISSION. 

455«  An  Agent,  Factor,  or  Broker,  is  a  person  who  trans- 
acts business  for  another. 

456.  A  Commission  Merchant  is  an  agent  .who  buys  and 
sells  goods  for  another. 

45 y.  Commission  is  the  f&e  or  compensation  of  an  agent, 
factor,  or  commission  merchant. 

458.  A  Consignment  is  a  qu«intity  of  goods  sent  to  one  person 
to  be  sold  on  commission  for  another  person. 

459.  A  Consignee  is  a  pei^ptn  who  receives  goods  to  sell  for 
another;  and 

46®.  A  Consignor  is  a  person  who  sends  goods  to  another  to 
be  sold. 

461*   The  Net  Proceeds  of  a  sale  or  collection  is  the  sum 

left,  ?ifter  deducting  the  commission  and  other  charges. 

NoTB. — A  person  who  is  employed  in  establishino^  mercantile  relations  between 
other.s  living  nt  a  distance  from  each  other,  is  called  the  Correspoudenf  of  the 
party  in  whose  behalf  he  acts.  A  correspondent  is  the  a</eut  of  those  whose 
custom  or  patronage  he  secures  to  the  party  in  whose  interest  he  is  employed. 

402.  Commission  is  usually  reckoned  at  a  certain  per  cent,  of 
the  money  involved  in  the  transaction ;  hence  we  have  the  follow- 
ing relations : 

I.  Commission  \^  percentage ^  (445). 

II.  The  sum  received  by  the  agent  as  the  price  of  property  sold, 
or  the  sum  invested  by  the  agent  in  the  purchase  or  exchange  of 
property,  is  the  hase  of  commission,  (446). 


COMMISSION. 


269 


I 


III,  The  sum  remitted  to  an  agent,  and  including  both  the  pur- 
chase money  and  the  agent's  commission,  is  the  amount,  (4:4:T). 

lY.  The  sum  due  the  employer  or  consignor  as  the  net  proceeds 
of  a  sale  or  collection,  is  the  difference,  (441:§). 

EXAMPLES    r®R   PRACTICE. 

1.  My  agent  sells  goods  to  the  amount  of  $6250;  what  is  his 
commission  at  3  ^  ? 

OPERATION.  Analysis.    According  to 

$6250  X  .03  =  $187.50  ^^^^-  ^'  (4^^)'  ^^^®  "^^1^^" 

ply  the  sum  obtained  for 

the  goods,  $0250,  which  is  the  base  of  the  commission,  (11),  by  the 
rate  of  the  commission,  .03,  and  obtain  the  commission  or  percent- 
age, $187.50. 

2.  A  flour  merchant  remits  to  his  agent  in  Chicago  $3796,  for 
the  purchase  of  grain,  after  deducting  the  commission  at  4  %  ; 
how  much  will  the  agent  expend  for  his  employer,  and  what  wil) 
be  his  commission  ? 

OPERATION.  Analysis.      Ac- 

1.00  -f  .04  =  1.04  cording     to    Prob. 

$3796  -^  1.04  =  $3650,  for  grain,  IV,  (452),  we   di- 

$3796  —  $3650  =  $146,  commission.  vide  the  remittance, 

$3796,  which  is 
amount,  (III),  by  1  plus  the  rate  of  commission,  or  1.04,  and  obtain 
the  base  of  commission,  $3050,  wk^ch  is  the  sum  to  be  expended  in 
the  purchase.  Subtracting  this  from  the  remittance,  we  have  $146, 
the  commission. 

NoTK  1. — It  is  evident  that  the  whol©  remittance,  $?.7D6,  should  not  he  tnken 
as  the  hase  of  commission  ;  for  that  vvoi^'l  be  computinir  commission  on  commis- 
sion. A  person  must  charge  commissioi  only  on  what  he  expends  or  collects,  in 
his  capacify  as  agent. 

3.  A  factor  sold  real  estate  on  commission  of  5  %,  and  returned 
to  the  owner,  as  the  net  proceeds,  $8075 ;  for  what  price  did  he 
sell  the  property,  and  what  was  his  commission  ? 

OPERATION.  Analysis.    According  to 

1  QO Q5  g^   95  Prob.  V,    (453),  we  divide 

$8075 -r- .95  =  $8500,  price.  the    net    proceeds,   $8075, 

$8500  —  $8075  =  $425,  com.         which  is  difference,   (IV), 

by   1   minus   the   rate  of 
23* 


270  PERCENTAGE, 

commission,  and  obtain  the  base,  $8500,  which  is  the  price  of  the 
property  sold  ;  whence  by  subtraction,  we  obtain  the  commission, 
$425. 

•  4.  An  agent  sold  my  house  and  lot  for  §8600 ;  wliat  was  his 
commission  at  2i  ^  ?  A71S.  $193.50. 

5.  A  lawyer  collects  8750.75;  what  is  his  commission  at 
3i  %?  "  Ans.  $28.15  +  . 

6.  My  agent  in  New  York  has  sold  3500  bushels  of  Indiana 
wheat  @  $1.40,  and  3600  bushels  of  dent  corn  @  $.74;  what  is 
his  commission  at  2}  %  ? 

7.  A  dealer  in  Philadelphia  sells  hides  on  commission  of  8}  %, 
as  follows:  2000  lb.  Orinoco  @  $.23 J,  5650  lb.  Central  Ameri- 
can  @  $.22,  450  lb.  Texas  @  $.23,  and  650  lb.  city  slaughter 
@  $.21;  wjiat  does  he  receive  for  hi;,  services?     Ans^.  $162.75. 

8.  A  commission  merchant  sold  a  consignment  of  flour  and  pork 
for  $25372.  He  charged  $132  for  storage,  and  6i  %  commis- 
sion ;  what  were  the  net  proceeds  of  the  sale  ? 

9.  An  agent  for  a  Rochester  nurseryman  sells  4000  apple  trees 
at  $25  per  hundred,  2000  pear  trees  at  $50  per  hundred,  1600 
peach  trees  at  $20  per  hundred,  1800  cherry  trees  at  $50  per 
hundred,  and  500  plum  trees  at  $50  per  hundred ;  what  is  his 
commission  at  30  %,  and  how  much  should  he  return  to  his  em- 
ployer as  the  net  proceeds,  after  deducting  $203.50  for  expenses? 

Ans.  Commission,  $1041 ;  Net  proceeds,  $2225.50. 

10.  A  lawyer  having  a  debt  of  $785  to  collect,  compromises 
for  82  %  ;  what  is  his  commission,  at  5  %  ?         Ans.  $32,185. 

11.  I  purchased  in  Chicago  4000  bushels  of  wheat  @  $1.25, 
and  shipped  the  same  to  my  agent  in  Oswego,  N.  Y.,  who  sold 
it  @  $1.50  ;  how  much  did  I  make,  after  paying  expenses  amount- 
ing to  $415,  and  a  commission  of  3  %  ?  Ans.  $405. 

12.  An  agent  received  $63  for  collecting  a  debt  of  $1260; 
what  was  the  rate  of  his  commission?  Ans.  5  ^. 

13.  My  Charleston  agent  has  charged  $74.25  for  purchasing 
26400  lb.  of  rice  at  $4.50  per  100  lb.;  required  the  rate  of  his 
commission.  ■{-. 

14.  A  house  and  lot  were  sold  for  $7850,  and  the  owner  re- 


COMMISSION. 


271 


ceived  $7732.25  as  the  net  proceeds;  what  was  the  rate  of  com- 
mission ?  '^Z  5" 

15.  A  commission  merchant  in  Boston  having  received  28000 
lb.  of  Mobile  cotton,  eiFects  a  sale  at  $.12 J  per  pound.  After 
deducting  $35. 3G  for  freight  and  cartage,  $10.50  for  storage,  and 
his  commission,  he  remits  to  his  employer  $3252.89  as  the  net 
proceeds  of  the  sale ;  at  what  rate  did  he  charge  commission  ? 

Ans.  5i  %. 

16.  The  net  proceeds  of  a  sale  were  $5635,  the  commission  was 
$115 ;  what  was  the  rate  of  commission  ? 

17.  An  agent  received  $22.40  for  selling  grain  at  a  commission 
of  4  ^c ;  what  was  the  value  of  the  grain  sold  ?  Ans.  5 GO, 

18.  My  attorney,  in  collecting  a  note  forme  at  a  commission  of 
8  ffcy  received  as  his  fee  $6.80  ;  what  was  the  face  of^he  note  ? 

19.  Sent  to  my  agent  in  Boston  $255,  to  be  invested  in  French 
prints  at  $.15  per  yard,  after  deducting  his  commission  of  2  %  3 
how  many  yards  shall  I  receive  ?  Ans.  1G6G!;. 

20.  John  Kennedy,  commission  merchant,  sells  for  Ladd  &  Co. 
860  barrels  of  flour  @  $7.50,  on  a  commission  of  2i  %.  lie 
invests  the  proceeds  in  dry  goods,  after  deducting  his  commission 
of  1  i  %  for  purchasing ,  how  many  dollars*  worth  of  goods  do 
Ladd  &  Co.  receive?  Ans.  $6195.81  +  . 

,  21.  A  commission  merchant,  whose  rate  both  for  selling  and 
investing  is  5  %,  receives  24000  lbs.  of  pork,  worth  6  cents,  and 
$3000  in  cash,  with  instructions  to  invest  in  a  shipment  of  cotton 
to  London.     What  will  be  his  entire  commission  ?     Ans.  $280. 

22.  A  speculator  received  $3290  as  the  net  proceeds  of  a  sale, 
after  allowing  a  commission  of  6  ^  ;  what  was  the  value  of  the 
property?  Ans.  ^3500. 

23.  The  net  proceeds  of  a  shipment  of  500  tons  of  pressed  hay, 
after  deducting  a  commission  of  3  % ,  and  $500  for  other  charges, 
were  $6290 ;  what  was  the  selling  price  per  ion  ?  ,  ~^^M 

24.  I  send  a  quantity  of  dry  goods  into  the  country  to  be  sold^ 
at  auction,  on  commission  of  9  %.     What  amount  of  geods  must 
be  sold,  that  my  agent  may  buy  produce  with  the  avails,  to  the 
Value  of  $3500,  after  retaining  his  purchase  commission  of  4  ^  ? 


272  PERCENTAQB 

Note  2.  —  $3500  plus  the  agent's  commission  equals  the  net  proceeds  of  the 
sale. 

25.  Having  sold  a  consignment  of  cotton  on  3  ^  commission, 
I  am  instructed  to  invest  the  proceeds  in  city  lots,  after  deducting 
my  purchase  commission  of  2  %.  My  whole  commission  is  §265; 
what  is  the  price  of  the  city  lots?  Ans.  $5141. 

26.  What  tax  must  be  assessed  to  raise  $50000,  the  collector's 
commission  being  J  %  ?  Ans.  $50377.83  +  . 

STOCKS. 

463.  A  Company  is  an  association  of  individuals  for  the 
prosecution  of  some  industrial  undertaking.  Companies  may  bo 
incorporated  or  unincorporated. 

41G4:.  A  Corporation  is  a  body  formed  and  authorised  by  law 
to  act  as  a  single  person. 

4:&3.  A  Charter  is  the  legal  act  of  incorporation,  and  defines 
the  powers  and  obligations  of  the  incorporated  body. 

4:00.   A  Firm  is  the  name  under  which  an  unincorporated 

company  transacts  business. 

Note. — A  private  banking  company,  or  a  manufacturing  or  commercial  firm 
\s  also  called  a  House. 

4:G7«  The  Capital  Stock  of  a  corporation  is  the  money  con- 
tributed and  employed  to  carry  on  the  business  of  the  company. 

468.  Joint  Stock  is  the  money  or  capital  of  any  company, 
incorporated  or  unincorporated. 

469.  Scrip  or  Certificates  of  Stock  are  the  papers  or  docu- 
ments issued  by  a  corporation,  giving  the  members  their  respective 
titles  or  claims  to  the  joint  capital. 

470.  A  Share  is  one  of  the  equal  parts  into  which  capital 
stock  is  divided.  The  value  of  a  share  in  the  original  contribu- 
tion of  capital  varies  in  different  companies;  in  bank,  insurance, 
and  railroad  companies  of  recent  erganization,  it  is  usually  $100. 

471.  Stockholders  are  the  o\^ners  of  stock,  either  by  original 
title  or  by  subsequent  purchase.  The  stockholders  constitute  the 
company. 

Notes. — 1.  The  capital  stock  of  any  corporation  is  limited  by  the  charter.  As 
a  general  rule,  only  a  portion  is  paid  at  the  time  of  subscription,  the  residue 
being  reserved  for  future  outlays  or  disbursements. 


STOCKS. 


273 


2.  When  the  cnpital  stock  has  been  all  paid  in,  money  maybe  raised,  if  neces- 
sary, by  luoim,  t^ecured  by  inortg:ige  ui)(>n  the  property.  The  OohcIh  ist^ued  lor 
tlje.<e  b.aiis  entitle  tl^e  holder:?  to  a  tixed  rale  of  inieresjt.     . 

o.  Stocks,  as  a  generiil  name,  applies  t<»  the  t'ciip  and  bonds  of  a  corporntion, 
to  goveiriMient  bonds  and  pnbiic  .vecuriiits,  and  to  .ill  paper  represeuiing  ji>iiit 
c;ipit;ii  or  elaiuis  upon  corporate  I'odies. 

4.  'J'he  members  of  an  incorporated  company  are  individually  liable  for  the 
debts  and  obligations  of  the  coin[»any,  to  the  amount  <if  tiieir  interest  or  stock 
in  the  company,  and  to  no  greater  amount.  But  the  members  of  a  firm  or  h<»use 
are  individually  liahle  f<<r  all  the  debts  and  obligations  of  the  company,  without 
regard  to  the  amount  of  their  share  or  interest  in  the  concern. 

The  calculations  of  percentage  in  stocks  are  treated  in  this  work 
under  the  heads  of 

Stock-jobbing,  Assessments  and  Dividends,  and  Stock  Invest- 
ments. 


STOCK-JOBBING. 

47S.  stock-jobbings  is  the  buying  and  selling  of  stocks  with 
a  view  to  realize  gain  from  their  rise  and  fall  in  the  market. 

47S.  The  Nominal  or  Par  value  of  stock  is  the  sum  for 
which  the  scrip  or  certificate  is  issued. 

474.  The  Market  or  Eeal  value  of  stock  is  the  sum  for 
which  it  will  sell. 

47o.  Stock  is  At  Par  when  it  sells  for  its  fir.st  cost,  cr 
nominal  value. 

47G.  Stock  is  Above  Par,  at  a  premium  or  advance,  when  it 
sells  for  more  than  its  nominal  value. 

477.  Stock  is  Below  Par,  or  at  a  discount,  when  it  sells  for 
less  than  its  nominal  value. 

NoTi^. — When  the  business  of  a  compjiny  pays  large  profits  to  ♦he  ctock- 
holders,  the  stock  will  be  worth  nH»re  thitn  iiy  original  cost;  but  when  the  nasi' 
ness  does  not  pay  expenses,  the  value  of  the  stock  will  be  less*  than  its  original 
cost.  The  averai^e  market  value  of  stock  generally  varies  directly  as  thti  rate 
of  profit  which  the  business  pays. 

478.  A  Stock  Broker  is  a  person  who  buys  and  sells  stacks, 
either  for  himself,  Or  as  the  agent:  of  another. 

NfeTK.— :-A  person  employed  by  a  manufacturer,  wholesale  dealer,  or  cornm^ssinn 
merchant,  to  seek  customers  and  close  bar<rains,  at  or  from  his  place  of  bi^ouieLS, 
is  called  a  broker,  of  the  class  or  kind  corresponding  to  his  bui^inci^s. 

479.  Brokerage  is  the  fee  or  compensation  of  a  brok*^.i-. 

480.  The  calculations  in  stock-jobbing  are  based  upon  the 
followino;  relations : 


274  PERCENTAGE. 

I.  Premium,  discount,  and  brokerage  are  each  a  percentage, 
computed  upon  the  par  value  of  the  stock  as  the  base. 

II.  The  market  value  of  stock,  or  the  proceeds  of  a  sale,  is  the 
amount  or  difference^  according  as  the  sum  is  greater  or  less  than 
the  par  value. 

NoTK  1. — In  .ill  examples  relating  to  stocks,  $100  will  be  considered  a  share, 
unless  otherwise  stated. 

EXAIttPLES    FOR   PRACTICE. 

1.  What  cost  54  shares  of  Reading  Railroad  stock,  at  4J  ^ 

premium  ? 

oPERATiox.  Analysis.      We    first 

S5400  X    .045  =  ^243,  premium.        compute    the     premium 

$5400  +  §243  =  5643,   An^.  upon  the  par  value  of  the 

Or,  stock,  and  find  it  to  be 

5400  X  $1,045  =  $5643,  An^,         $243  ;  adding  this  to  the 

$5400,  we  obtain  the  cost, 
or  market  value,  $5G43.  Or,  since  every  dollar  of  the  stock  will  cost 
$1  plus  the  premium,  or  $1,045,  $5400  will  cost  5400  X  $1,045  = 
$5G43. 

2.  What  do  I  receive  for  32  shares  of  telegraph  stock,  which  a 
broker  sells  for  me  at  15  %  discount  charging  }  %  brokerage? 

OPERATION.  Analysis.        Adding 

15  _(_      00'^ 5  -^      15'^5  the  rate  of  brokerage  to 

^LOO  —  $;i525  =  $.8475  proceeds  """^^  ^^^^  ^^  discount,  we 

of  $1  of  stock.  ^^a^^e    4525  ;    hence   $1 

3200  X  $.8475  =  §2712,  Am.  ^^iH  bring  $1— $.1525= 

$.8475,  and  $3200  will 
bring  3200  X  $.8475  =  $2712. 

3.  I  put  §35400  into  the  hands  of  a  broker  to  be  invested  in 
Missouri  State  Bonds,  when  their  market  value  is  12  %  below  par; 
how  many  shares  shall  I  receive,  if  the  broker  charges  J  %  for 
his  services? 

OPERATION. 

$1.00  —  $.12    =^  $.88,  market  value  of  $1. 
$  .83  +  $.00.i  =r  .885,  cost  of  $1. 
§35400  ~  .885    =  §40000  =  400  shares,  Ans. 
Analysis.     Since  the  stock  is  12  %  below  par,  the  market  value  of 
$1  is  $.88  ;  adding  the  rate  of  brokerage,  we  find  that  every  dollar  of 


h 


/     i  /^ 

STOCKS.  .  2^5 

the  stock  will  cost  me  $.885.   Hence  for  $35400  the  broker  can  buy 

$35400  —  .885  =  $40000  =  400  shares. 

NoTKS.  — 2.  The  rate  of  brokerage  in  New  York  city  has  been  fixed  by  cus- 
tom at  \  per  cent. 

3.  Since  brokerage  has  the  same  base  as  the  premium  or  (Hscount,  the  rute  of 
brokerage  may  always  be  combined  with  the  rate  of  premium  or  discount,  by 
addition  or  w'^ubtraction,  as  the  nature  of  the  quej^tion  may  require. 

4.  The  price  of  stock  is  usually  quoted  at  a  certain  per  cent,  of  the  fnce,  or 
nominal  value.  Thus  stock  at  4  ^  above  par  is  quoted  at  104  ^  \  slock  at  5  ^jo 
below  par  is  quoted  at  95  %  ;  and  so  on. 

4.  What  is  the  market  value  of  15  Ohio  State  bonds  at  112  %  ? 

AuA.  §1680. 

5.  What  shall  I  realize  on  20  shares  of  Panama  railroad  stock 
at  135  %,  brokerage  at  If  %  ?  .4»,s.  $2665. 

6.  My  agent  bought  for  me  120  shares  of  N".  Y.  Central  rail- 
road stock,  paying  80 J  %,  and  charging  brokerage  at  A  %  ;  what 
did  the  stock  cost  me?  An^.  $9750. 

7.  What  cost  36  shares  in  the  Merchants*  Bank,  at  a  premium 
of  7J  %,  brokerage  i  %  ? 

8.  A  speculator  invested  $21910  in  shares  of  the  Harlem  rail- 
road, at  a  discount  of  60 J  %  \  how  many  shares  did  he  buy  ? 

9.  If  400  shares  of  the  Bank  of  Commerce  sell  for  §40150, 
what  is  the  rate  of  premium?  Am.  |  ^. 

10.  A  broker  receives  §48447  to  be  invested  in  bonds  of  the 
Michigan  Central  railroad,  at  94 1  %  ;  how  much  stock  can  he 
buy,  allowing  1 }  %  brokerage  ? 

11.  My  agent  sells  830  barrels  of  Genesee  flour  at  §6  per  barrel, 
commission  5  %,  and  invests  the  proceeds  in  stock  of  the  Penn- 
sylvania Coal  Company,  at  82 J  %,  charging  }  %  for  making  the 
purchase;  how  many  shares  do  I  receive  ?  Ans.  57. 

12.  I  purchased  18  shares  of  Ocean  Telegraph  stock,  par  value 
§500  per  share,  at  a  premium  of  2  %,  and  sold  tlie  same  at  a  dis- 
count of  28  %  ;  what  was  my  loss  ?  Am.  §2700. 

Note  5.—  The  rate  of  loss  is  .02.  +  .28  =»  .30,  or  .30  %. 

13.  A  speculator  exchanged  $3600  of  railroad  bonds,  at  5  % 
discount,  for  27  shares  of  stock  of  the  Suffolk  Bank,  at  3  % 
premium,  receiving  the  difference  in  cash;  how  much  money  did 
he  receive? 

14.  A  merchant  owning  525  shares  in  the  American  Exchange 


276  *      PERCENTAGE. 

Bank,  worth  104   %,  exchanges   them  for  United  States  bonds 
worth  105  %  ;  how  much  of  the  latter  stock  does  he  receive? 

15.  I  purchased  12  shares  of  stock  at  a  premium  of  5  %,  and 
sold  the  same  at  a  loss  of  396 ;  what  was  the  selling  price  ? 

16.  Having  bought  $64000  stock  in  the  Cunard  Line,  at  2  % 
premium,  at  what  price  must  I  sell  it,  to  gain  $2560  ? 

Ans.  106  %. 

17.  A  speculator  bought  250  shares  in  a  Carson  Valley  mining 
company  at  103  %,  and  150  shares  of  the  Western  Railroad  stock 
at  95  ^0  '}  ^^  exchanged  the  whole  at  the  same  rates,  for  shares  in 
the  N.  Y.  Central  Railroad  at  80  %,  which  he  afterward  sold  at 
85  %.     How  much  did  he  gain?  Ans.  $2500. 

18.  I  purchased  stock  at  par,  and  sold  the  same  at  3  %  pre- 
mium, thereby  gaining  $750;  how  many  shares  did  I  purchase? 

19.  A  broker  bought  Illinois  State  bonds  at  103  %,  and  sold 
at  105  %.  His  profits  were  $240;  what  was  the  amount  of  his 
purchase?  Ans.  $12000. 

20.  A  man  invested  in  mining  stock  when  it  was  4  %  above 
par,  and  afterward  sold  his  shares  at  5  J  %  discount.  His  loss 
in  trade  was  $760;  how  many  shares  did  he  purcbp&se  ? 

21.  I  invested  $6864  m  Government  bonds  at  106  %,  paying 
1}  (fo  brokerage,  and  afterward  sold  the  stock  at  112  %,  paying 
IJ  %  brokerage;  what  was  my  gain  ?  Ans.  $208. 

22.  How  much  money  must  be  invested  in  stocks  at  3  ^  ad- 
vance, in  order  to  gain  $480  by  selling  at  7  %  advance  ? 

23.  How  many  shares  of  stock  must  be  sold  at  4  %  discount, 
lirokerage  i  %,  to  realize  $4775?  Ans.  50. 

INSTALLMENTS,    ASSESSMENTS,    AND    DIVIDENDS. 

4(:8l«  An  Installment  is  a  portion  of  the  capital  stock  re: 
quired  of  the  stockholders,  as  a  payment  on  their  subscription,    y. 

482.  An  Assessment  is  a  sum  required  of  stockholders,  to 
meet  the  losses  or  the  business  expenses  of  the  company. 

48 3«  A  Dividend  is  a  sum  paid  to  the  stockholders  from  the 
profits  of  the  business. 


STOCKS. 


277 


484.  Gross  Earnings  are  all  the  moneys  received  from  the 
regular  business  of  the  company. 

485.  K'et  Earnings  are  the  moneys  left  after  paying  expenses, 
losses,  and  the  interest  upon  the  bonds,  if  there  be  any. 

48G.  In  the  division  of  the  net  earnings,  or  the  apportion- 
ment of  dividends  and  assessments,  the  calculations  are  made  by 
finding  the  rate  per  cent,  which  the  sum  to  be  distributed  or  as- 
sessed bears  to  the  entire  capital  stock.     Hence^ 

487.  Dividends  and  assessments  are  a  percentage  computed 
upon  the  par  value  of  the  stock  as  the  base. 

EXAMPLES    FOR    PRACTICE. 

1.  The  Long  Island  Insurance  Company  declares  a  dividend 
of  6  %  ;  what  does  A  receive,  who  owns  14  shares  ? 

Analysis.       According     to 

OPERATION.  449,    we    multiply    the    base, 

$1400  X  .06  =  $84      -        $1400,   by  the  rate,  .06,  and 

obtain  the  dividend,  $84. 

2.  A  canal  company  whose  subscribed  funds  amount  to  $84000, 
requires  an  installment  of  $6300;  what  per  cent,  must  the  stock- 
holders pay? 

OPERATION.  Analysis.    According  to 

$8300  ~  84000  =  .07i  450,    we   divide    the    in- 

stalhnent,  $6300,  which  is 
percentage,  by  the  base,  $84000,  and  obtain  the  rate,  .07 J  =  7J  %, 

3.  A  man  owns  56  shares  of  railroad  stock,  and  the  company 
has  declared  a  dividend  of  8  %  ;  what  does  he  receive  ? 

Ans.  $448. 

4.  I  own  $15000  in  a  mutual  insurance  company;  how  many 
shares  shall  I  possess  after  a  dividend  of  6  %  has  been  declared, 
payable  in  stock  ?  Ans.   159  shares. 

5.  The  Pittsburgh  Gas  Company  declares  a  dividend  of  15  %  ; 
what  will  be  received  on  65  shares  ? 

6.  A  received  $600  from  a  4  %  dividend ;  how  much  stock 
did  he  own  ?  '  Ans.  $15000. 

24 


278  PERCENTAGE. 

7.  The  paid-in  capital  of  an  insurance  company  is  $536000. 
Its  receipts  for  one  year  are  $99280,  and  its  losses  and  expenses 
are  $56400;  w^at  rate  of  dividend  can  it  declare?     Ans.  8  %. 

8.  The  net  earnings  of  a  western  turnpike  are  $3616,  and  the 
amount  of  stock  is  $56000 ;  if  the  company  declare  a  dividend 
of  6  % ,  what  surplus  revenue  will  it  have  ?  Ans.  $256. 

9.  The  capital  stock  of  the  Boston  and  Lowell  Railroad  Co. 
is  $1830000,  and  its  debt  is  $450000.  Its  gross  earnings  for  the 
year  1858  were  $407399,  and  its  expenses  $217621.  If  the  com- 
pany paid  expenses,  and  interest  on  its  debt  at  5|  %,  and  reserved 
$78,  what  dividend  would  a  stockholder  receive  who  owned  30 
shares?  Ans.  $270. 

10.  The  charter  of  a  new  railroad  company  limits  the  stock  to 
$800,000,  of  which  3  instalhnents  of  10  %,  25  %,  and  35  %,  re- 
spectively, have  been  already  paid  in.  The  expenditures  in  the  con- 
tstruction  of  the  road  have  reached  the  sum  of  $540,000,  and  the 
estimated  cost  of  completion  is  $400,000.  If  the  company  call  in 
the  final  installment  of  its  stock,  and  assess  the  stockholders  for  tho 
remaining  outlay,  what  will  be  the  rate  %  ?  Ans.  17  J. 

11.  The  Bank  of  New  York,  having  $156753.19  to  distribute 
to  the  stockholders,  declares  a  dividend  of  bi  %  ;  what  is  the 
amount  of  its  capital  ?  Ans.  $2,985,775  nearly. 

12.  The  passenger  earnings  of  a  western  railroad  in  one  year 
were  $574375.25,  the  freight  and  mail  earnings  were  $643672.36, 
the  whole  amount  of  disbursements  were  $651113.53,  and  the 
>^ompany  was  able  to  declare  a  dividend  of  8  %  ;  how  much  scrip 
had  the  company  issued?  Ans.  $7086676. 

13.  Having  received  a  stock  dividend  of  5  %,  I  find  that  I 
own  504  shares ;  how  many  shares  had  I  at  first  ?       Ans.  480. 

14.  I  received  a  6  %  dividend  on  Philadelphia  City  railroad 
stock,  ajid  invested  the  money  in  the  same  stock  at  75  %.  Mj 
stock  had  then  increased  to  $16200;  what  was  the  amount  of  my 
dividend  ?--'\  ]--  1  ^' '         ;     vA  ^  f   5    ^  '  ^ '    Ans.  $900. 

.15.  A  ferry  company,  whose  stock  is  $28000,  pays  5  %  divi- 
dends semi-annually..  The  annual  expenses  of  the  ferry  are 
$2050 ;  what  are  the  gross  earnings  ?  Ans.  $5750. 


STOCKS.  279 

STOCK  IXTEST:\rEXTS.* 

4:SS.  The  net  earnings  of  a  corporation  are  usually  divided 
among  the  stockholders,  in  semi-annual  dividends.  The  income  of 
capital  stock  is  therefore  fluctuating,  being  dependent  upon  the  con- 
dition of  business ;  while  the  income  arising  from  bonds,  whether  of 
government  or  corporations,  is  fixed,  being  a  certain  rate  per  cent., 
annually,  of  the  par  value,  or  face  of  tlie  bonds. 

489.  Federal  or  United  States  Securities  are  of  two  kinds : 
viz..  Bonds  and  Notes. 

Bonds  are  of  two  kinds. 

First,  Those  which  are  payable  at  a  fixed  date,  and  are  known 
and  quoted  in  commercial  transactions  by  the  rate  of  interest  they 
bear,  thus,  :  U.  S.  6's,  that  is,  United  States  Bonds  bearing  6  % 
interest. 

Second,  Those  which  are  payable  at  a  fixed  date,  but  which  may 
be  paid  at  an  earlier  specified  time,  as  the  Government  may  elect. 
These  are  known  and  quoted  in  commercial  transactions  by  a  cond)i- 
nation  of  the  two  dates,  thus :  U.  S.  5-20's,  or  a  combination  of  the 
rate  of  interest  and  the  two  dates,  thus :  U.  S.  6's  5-20 ;  that  is, 
bonds  bearing  G  %  interest,  which  are  payable  in  twenty  years,  bub 
may  be  paid  in  five  years,  if  the  Government  so  elect. 

When  it  is  necessary,  in  any  transaction,  to  distinguish  from  each 
other  different  issues  which  bear  the  same  rate  of  interest,  this  is 
done  by  adding  the  year  in  which  they  become  due,  thus  :  U.  S.  5's 
of  71 ;  U.  S.  5's  of  74 ;  U.  S.  6's  5-20  of  '84 ;  U.  S.  6's  5-20 
of  '85. 

Notes  are  of  two  kinds. 

First,  Those  payable  on  demand,  without  interest,  known  as  United 
States  Legal-tender  Notes,  or,  in  common  language,  "Green  Backs." 

*  TJie  following  eight  pages  contain/owr  pages  of  new  matter,  on  U.S.  Secniitics, 
Bonds,  Treasury  Notes,  Gold  Investments,  &c.,  to  meet  a  necessity  which  did  not 
exist  ut  tlie  time  this  book  was  written. 

The  pupil  will  find  the  Cases,  Rules,  and  Operations  of  the  previous  editions 
essentially  the  same  in  this,  with  additional  examples,  and  other  matter,  which  may 
be  used  or  omitted;  so  that  the  present  may  be  used  with  the  preyious  editions 
fuith  little  or  no  inconvenience. 


280  PERCENTAGE. 

Second^  Notes  payable  at  a  specified  time,  with  interest,  known 
as  Treasury  Notes.  Of  these,  there  are  two  kinds,  —  Six-per-cent. 
Compound-interest  Notes,  and  Notes  bearing  7i\  %  interest,  tlie 
latter  known  and  quoted  in  commercial  transactions  as  T.CO's. 

The  nomenclature  here  explained  is  the  one  used  in  commercial 
transactions,  which  involve  similar  securities  of  States  or  corporations. 

The  interest  on  all  bonds  is  payable  in  gold. 

The  interest  on  notes  is  payable  in  Legal-tender  Notes. 

When  Bonds  or  Stocks  are  sold,  a  revenue  stamp  must  be  used 
equal  in  value  to  one  cent  on  each  $100,  or  fraction  of  $100,  of  their 
currency  value.  If  sold  by  a  broker,  this  is  charged  to  the  person 
for  w^hom  they  are  sold. 

The  following  are  the  principal  United  States  Securities :  — 

BONDS. 

U.  S.  G's  of  1867. 

U.  S.  G's  of  18G8. 

U.  S.  G's  of  1880. 

U.  S.  G's  of  1881. 

U.  S.  5's  of  1871. 

U.  S.  5's  of  1874. 

U.  S.  6-20's,  due  in  1882,  interest  5  %. 

U.  S.  5-20's,  due  in  1884,  interest  G  %. 

U.  S.  5-20's,  due  in  1885,  interest  G  %. 

U.  S.  10-40's,  due  in  1004,  interest  5  %. 

Pacific  Railroad  G's  of  1895. 

Pacific  Ptailroad  G's  of  1896. 

NOTES. 

Compound-interest  Notes  of  18G7. 
Compound-interest  Notes  of  1868. 
7.30  Notes  of '1867. 
7.30  Notes  of  1868. 


STOCKS.  281 

CASE   I. 

490.  To  find  what  income  any  investment  will  pro- 
duce. 

1.  What  income  will  be  obtained  by  investing  $G840  in  stock 
bearing  6  % ,  and  purchased  at  95  %  ? 

OPERATION.  Analysis.    We  di- 

$8840  -  .95  =  $7200,  stock  purchased,     l'^^'^?    i-^^^^^"^; 

$7200  X  .06  =  $432,  annual  income.  ^^^^^'  \^^^  ^^^*  ^^ 

Si,  and  obtain  $7200, 

the  stock  which  the  investment  will  purchase,  (452).  And  since  the 
stock  bears  6  %  interest,  we  have  $7200  X  -OG  ==  $432,  the  annual 
income  obtained  by  the  investment.     Hence, 

Rule.  — Fmd  how  much  stock  the  investment  will  purchase,  and 
then  compute  the  income  at  the  given  rate  upon  the  par  value, 

EXAMPLES   FOR    PRACTICE. 

1.  The  trustees  of  a  school  invested  $35374.80  in  the  U.  S. 
5  %  bonds  as  a  teachers'  fund,  purchasing  the  stock  at  102J^ 

if  the  salary  of  the  Principal  be  $1000,  what  sum  will  be  left  to  pay 
assistants?  Ans.  $725. CO. 

2.  A  young  man,  receiving  a  legacy  of  $48000,  invested  one  half 
in  5  %  stock  at  95  J  %,  and  the  other  half  in  G  %  stock  at  112  %, 
paying  brokerage  at  J  %  ;  what  annual  income  did  he  secure  from 
his  legacy?  Ans.  $2530. 

3.  I  have  S2300  to  invest,  and  can  buy  New  York  Central  6's 
at  85  %,  or  N^w  York  Central  7 's  at  95  %  ;  how  much  more  prof 
itable  will  tlj9  latter  be  than  the  former,  per  year  ? 

4.  A  owns  a  farm  which  rents  for  $411.45  per  annum.  If  he 
sell  the  same  for  $8229,  and  invest  the  proceeds  in  U.  S.  5-20 's  of 
'84,  at  105  %,  paying  \  %  brokerage,  will  his  yearly  income  be 
increased  or  duninished,  and  how  much  !     Ans.  Increased  $56.55. 

5.  A  sold  $8700  of  U.  S.  5-20's  of  '84  at  104  %,  paying  for 
necessary  revenue  stamps,  and  invested  the  proceeds  in  U.  S.  10-4 O's 
at  94  % ,  brokerage  \  %  both  for  selling  and  buying.  Did  he  gain 
or  lose  by  the  exchange,  and  how  much  annually? 

Ans.  $45.62—. 


282  PERCENTAGE. 

CASE  II. 

491.  To  find  what  sum  must  be  invested  to  obtain  a 
given  income. 

I.  What  sum  must  be  invested  in  Virginia  5  per  cent,  bonds, 
purchasable  at  80  % ,  to  obtain  an  income  of  $600  ? 

OPERATION.  Analysis.    Since 

$600  -~  .05  =  $12000,  stock  required.         ^^  ^^'  ^^^  stock  will 

$1200  X  .80  =z  $9600,  cost  or  investment,     obtain  S.05  income, 

to  obtain  SGOO  will 
require  $G00 -^.05  =  $12000,  (Case  1).  Multiplying  the  par  value 
of  the  stocks  by  the  market  price  of  $1,  we  have  $12000  X  -80  = 
S9600,  the  cost  of  the  required  stock,  or  tho  sum  to  be  invested. 
Hence  the 

IluLE.  I.  Divide  the  given  income  hy  the  %  which  the  stock 
pays  ;  the  quotient  will  he  the  par  value  of  the  stock  required. 

II.  Multiply  the  par  value  of  the  stock  hy  the  market  value  of 
one  dollar  of  the  stock  /  the  product  will  he  the  required  investment, 

EXAMPLES    FOR   PRACTICE. 

1.  If  Missouri  State  O's  are  16  %  below  par,  what  sum  must  bo 
invested  in  this  stock  to  obtain  an  income  of  $960  ? 

2.  What  sum  must  I  invest  in  U.  S.  5-20's  of  '82  at  962  %» 
brokerage  ^-  %,  to  secure  an  annual  income  of  $1500. 

Ans.  $29100. 

3.  How  much  must  I  invest  in  U.  S.  7-30's,  at  106  %,  that  my 
annual  income  may  be  $1752?  Ans.  $25440. 

4.  If  I  sell  $15600  U.  S.  10-40's  at  97  %,  and  invest  a  suf- 
ficient  amount  of  the  proceeds  in  U.  S.  5-20's  of  '85  at  107  %  to 
yield  an  annual  income  of  $540,  and  buy  a  house  with  the  re- 
mainder, how  much  will  the  house  cost  me  ?  Ans.  $5502. 

5.  Charles  C.  Thomson,  through  his  broker,  invested  a  certain 
sum  of  money  in  U.  S.  6's  5-20  at  107  %,  and  twice  as  much  in 
U.  S.  10-40's  at  98 J  %,  brokerage  in  each  case  I  %.  His  in- 
come from  both  investments  was  $1674.  How  much  did  ho  invest 
in  each  kind  of  stock  ? 

Ans.  First  kind,  $10692.     Second  kind,  $21384. 


STOCKS.  283 


CASE   III. 

492.  To  find  what  per  cent,  the  income  is  of  the  in- 
vestment, when  stock  is  purchased  at  a  given  price. 

1.  What  per  cent,  of  mj  investment  shall  I  secure  by  purchasing 
the  New  York  7  per  cents,  at  105  %  ? 

Analysis.     Since  Si  of  the  stock  . 
OPERATION.  will  cost  $1.05,  and  pay  S.07,  the  in- 

.07  -r-  1.05  =  Gf  %.  come  is  ^l^  =  6f  %  of  the  invest- 

ment.    Hence  the 

PtULE.  Divide  the  annual  rate  of  mcovfie  which  the  stock  hears 
hy  the  price  of  the  stock  ;  the  quotient  will  he  the  rate  vpon  the  in- 
vestment. 

EXAMPLES    rOPv   PIIACTICE. 

1.  What  per  cent,  of  his  money  will  a  man  obtain  by  investing 
in  G  per  cent,  stock  at  108  %  ?  Ans.  5f  %. 

2.  What  is  the  rate  of  income  upon  money  invested  in  G  per  cent, 
bonds,  purchased  at  a  discount  of  IG  %  ?  Ans.  7-J-  %. 

3.  Panama  railroad  stock  is  at  a  premium  of  34|^  %,  and  the 
charge  for  brokerage  is  IJ  %  ;  what  will  be  the  rate  of  income  on 
an  investment  in  these  funds  if  the  stock  pays  a  dividend  of  %\  % 
annually?  Ans.  6 J  %. 

4.  Which  is  the  better  investment,  to  buy  5's  at  70  %,  or  G's 
at  80  %  ? 

5.  Which  is  the  more  profitable,  to  buy  8's  at  120  %,  or  5's  a.t 
75%? 

G.  What  is  the  rate  of  income  upon  money  invested  in  U.  S. 
7-30's  at  lOG  %  ?  Ans.  GfJ  %. 

7.  Which  is  the  better  investment,  U.  S.  5-20's  of  '8-1  at 
108^  %,  or  U.  S.  10-40's  at  98  %,  and  how  much  per  cent,  per 
annum?  '  Ans.  U.  S.  5-20's,  -i%VV  %• 

8.  If  a  man  invest  $10000. in  U.  S.  10-40's  at  08  %,  and  ex- 
changes them  at  par  for  U.  S.  7-30's  at  102  %,  what  is  his  rate  of 
income?  ^ 

9.  What  per  cent  of  his  money  will  a  man  gam  by  investing  in 
Pacific  Ptailroad  G's  at  105  %  ? 


284:  PERCENTAGE. 

CASE  IV. 

493.  To  find  the  price  at  which  stock  must  be  pur- 
chased to  obtain  a  given  rate  upon  the  investment. 

1.  At  -wbat  price  must  6  per  cent,  stocks  he  purchased  in  order 
to  obtain  8  %  income  on  the  investment  ? 

OPERATION.  Analysis.      Since  $.06,  the  in- 

itio Ar>  AQ        c>^^  <^ome  of  SI  of  the  stock,  is  8  ^  of 

5).UO  -r-  .UO  =   Jt?iO.  .  •  1    />       -^  1  ^AAn\ 

the  sum  paid  lor  it,  we  have,  (449), 
$.06  -i-  .08  =  $75,  the  purchase  price.     Hence, 

HuLE.  Divide  the  annual  rate  of  income  which  the  stock  hears 
hy  the  rate  required  on  the  investment ;  the  quotient  will  he  the 
price  of  the  stock, 

EXAMPLES    FOR    PRACTICE. 

1.  What  must  I  pay  for  Government  5  per  cents.,  that  my  in- 
vestment may  yield  8  %  ?  Ans.  C2i-  %. 

2.  At  what  rate  of  discount  must  the  Vermont  G  per  cent,  bonds 
he  purchased  that  the  person  investing  may  secure  C^  %  upon  his 
money?  Ans,  4  %. 

3.  What  rate  of  premium  does  7  per  cent,  stock  bear  in  the  mar- 
ket when  an  investment  pays  G  %  ? 

4.  A  speculator  invested  in  a  Life  Insurance  Company,  and  re- 
ceived a  dividend  of  G  %,  which  was  8J^  %  on  his  investment;  at 
wbat  price  did  he  purchase?  Ans,  72  %. 

5.  What  must  I  pay  for  U.  S.  10-40's,  that  my  investment  may 
yield  6  %?  Ans,  83^  %. 

G.  What  rate  of  premium  docs  U.  S.  G's  5-20  bear  in  market 
when  an  investment  pays  5  %  ? 

7.  At  wbat  rate  of  discount  must  U  S.  7-30's  be  purchased, 
that  the  investment  sball  yield  10  %  ? 

8.  What  must  I  pay  for  government  G's  of  '81,  that  my  invest- 
ment may  yield  7  %  ? 


I 


STOCKS.  285 

GOLD  INVESTMENTS. 

493  a*  Currency  is  a  term  used  in  commercial  language,  First, 
To  denote  the  aggregate  of  Specie  and  Bills  of  Exchange,  Bank 
Bills,  Treasury  Notes,  and  other  substitutes  for  money  employed  in 
buying,  selling,  and  carrying  on  exchange  of  commodities  between 
various  nations.  Second,  To  denote  whatever  circulating  medium  is 
used  in  any  country  as  a  substitute  for  the  government  standard.  In 
this  latter  sense,  the  paper  circulating  medium,  when  below  par,  is 
called  Currency,  to  distinguish  it  from  gold  and  silver.  If,  from  any 
cause,  the  paper  medium  depreciates  in  value,  as  it  has  done  in  the 
United  States,  gold  becomes  an  object  of  investment,  the  same  as 
stocks.  In  commercial  language,  gold  is  represented  as  rising  and 
falling  ;  but  gold  being  the  standard  of  value,  it  cannot  vary.  The 
variation  is  in  the  medium  of  circulation  substituted  for  gold ;  hence, 
when  gold  is  said  to  be  at  a  premium,  the  currency,  or  circulating 
medium,  is  made  the  standard,  while  it  is  virtually  bolow  par. 

CASE    I. 

To  change  gold  into  currency. 

1.  How  much  currency  can  be  bought  for  $150  in  gold  when 
gold  is  at  170  %  ? 

OPERATION.  Analysis.     Since  a  dollar  of  gold  is 

<aji  *7(\  Ky  1  -A       <2?op:f^  worth  $1.70  in  currency,  there  can  be 

as  many  times  Si  70  of  currency  bought 

as  there  are  dollars  of  gold.     Therefore,  $1.70  X  150  =  $255  is  the 

amount  of  currency  which  can  be  purchased  for  $150  in  gold. 

BuLE.  Multiply  the  value  of  one  dollar  of  gold  in  currency  by 
the  number  of  dollars  of  gold, 

2.  What  is  the  value,  in  current  funds,  of  $250.47  gold,  when 
gold  is  at  1425  %  ?  Ans,  $357,546  — . 

3.  If  a  person  holds  $6000  U.  S.  10-40's,  what  would  be  his  an- 
nual income  in  current  funds  if  gold  is  at  157  %  ?      Ans.  $471. 

4.  A  merchant  purchased  a  bill  of  goods  for  which  he  was  to  pay 
$7000  in  currency,  or  $5500  in  gold,  at  his  option.  Will  he  gain 
or  lose  by  accepting  the  latter  proposition,  gold  being  at  138 J  %, 
and  how  much  in  currency?  Ans.  Lose  $617.50. 


9fi(^ 


PERCENTAGE. 


5.  Bought  broadcloth  @  $3  in  gold,  and  sold  the  same  @  $4  in 
currency.  Did  I  gain  or  lose  by  the  transaction,  and  how  much  per 
cent,  in  currency,  gold  being  at  140  %  ?  Ajis.  Lost  4Jf  %. 

C.    A  broker  invested  $3000  of  gold  in  U.  S.  G's,  vflnch.  were 
worth  102  %  in  currency.     What  was  his  annual  income  from  the 
investment,  gold  being  at  134  %  ?  and  what  the  rate  per  cent.  ? 
Ans.  to  frsf,  $236.47.     Ans.  to  last,  5|f  %. 

7.  A  gentleman  invested  $10,000,  current  funds,  in  U.  S.  5-20's 
of  '85,  at  104  %.  What  will  be  his  annual  income  in  currency 
when  gold  is  at  137  %  ?  Ans.  $790.38xV 

CASE   II. 

To  change  currency  into  gold. 

1.  How  much  gold  cau  be  purchased  for  $75  current  funds,  gold 
being  at  150  %  ? 

Analysis.     A  dollar  of  gold  cost  $1.50 

OPERATION.  in   currency,   therefore   there   can    be    as 

g'~5  -f-  $1.50  rz:  50  many  dollars  of  gold  purchased  for  $75  in 

currency  as  $1.50  is  contained  times  in  $75. 

HuLE.     Divide,  ike  amount  in  currency  hy  the  price  of  gold. 

2.  What  is  the  value  in  gold  of  a  dollar  in  currency  when  gold 
is  at  203  %  ?  Ans.  $.49/^^^. 

3.  Gold  being  the  standard,  what  is  the  rate  of  discount  upon  cur- 
rent funds  when  gold  is  at  134  %,  150  %,  175  %,  180  %,  215  %, 
and  398  %  ?      Ans.  to  first,  25ff  %.     Ans.  to  last,  74}|J-  %. 

4.  What  is  gold  quoted  at  when  a  dollar  in  currency  is  worth  20 
cents  in  gold,  30  cts.,  50  cts.,  15  cts.,  25  cts.,  and  72  cts.  ? 

5.  How  many  yards  of  cotton,  at  25  cts.  in  gold,  can  be  purchased 
for  $250  current  funds,  when  gold  is  at  175  %  ? 

Ans.  571?  yds. 

G.  Sold  $7800  7.30  Treasury  Notes,  at  105  %,  and  invested  the 
proceeds  in  gold  at  145  %,  with  which  I  bought  U.  S.  10-40\s  at 
GO  %  in  gold.  Will  my  yearly  income  be  increased  or  diminished 
by  the  transaction,  and  how  much  in  gold?      Ans.  Increased  $78. 

7.  Which  is  the  better  investment,  a  bond  and  mortgage  at  7  %, 
or  U.  S.  10-40's,  gold  being  134  ;  and  what  per  cent,  in  gold  ? 


I 


PROFIT  AND  LOSS.  287 

PROFIT   AND   LOSS. 

494:.  Profit  and  Loss  are  commercial  terms/used  to  express 
the  liaiii  or  loss  in  business  transactions. 

4LD*5.  Gains  and  losses  are  usually  estimated  at  some  rate  per 
cent,  on  the  money  first  expended  or  invested.      Hence 

I.  Profit  and  loss  are  reckoned  diS  percentage  upon  the  prime  or 
first  cost  of  the  goods  as  the  hase. 

II.  The  selling  price  of  the  goods  is  amount  or  difference,  ac- 
cording as  it  is  greater  or  less  than  the  prime  cost. 

EXAMPLES    FOR   PRACTICE. 

1.  A  merchant  bought  cloth  for  $3.25  per  yard,  and  gained 
8  %  in  selling ;  what  was  the  selling  price  ? 

OPERATION.  Analysis.    Multi- 

$3.25  X  .08  =  $.26,  advance  in  price.  P^yi'^g     the     prime 

$3.25  +  .26  =  $3.51,  selling  price.  cost,  $3.25,  which  is 

Qj.  the    hase    of    gain, 

$3.25  X  1.08  =  $3.51,  selling  price.  ^^j'    ^^   *''«   I"'"' 

.08,    we   have   $.26, 

the  gain,  which  added  to  the  cost  gives  $3.51,  the  selling  price.     Or, 

since  the  rate  of  gain  is  8  %,  that  which  cost  $1  will  bring  $1.08, 

and  the  selling  price  will  be  1.08  times  the  buying  price.     Hence 

$3.25  X  1.08  =  $3.51,  the  selling  price. 

2.  A  jobber  invested  $2560  in  dry  goods,  and  realized  $384 
net  profit ;  what  was  the  rate  per  cent,  of  his  gain  ? 

OPERATION.  Analysis.     According  to 

§38-1  -^  $2560  =  15  %  (Prob.  II,  450),  we  divide  the 

gain,  $384,  which  is  percent- 
age, by  the  cost,  $2560,  which  is  the  hase,  and  obtain  15  =  15  % ,  the 
rate  of  gain. 

3.  A  produce  dealer  sold  a  shipment  of  wheat  at  a  loss  of  5  ^, 
realizing  as  the  net  proceeds,  $8170;  what  was  the  cost? 

oPERATiow.  Analysis.     According  to 

$1.00 —  .05  =  .95  (Prob.V,  453),  we  divide  the 

8170  ^  .95  =  $8600,  An$.  net  proceeds,  $8170,  which 

is  diffei-ence,  (448),  by  1 
minus  the  rate  of  loss,  or  .95,  and  obtain  the  hase^  or  prime  cost,  $8600, 


i 


288  PERCENTAGE. 

4.  A  mercliant  pays  $7650  for  a  stock  of  spring  goods ;  if  he 
sell  at  an  advance  of  20  %  upon  the  purchase  price,  what  will  be 
his  profits,  after  deducting  $480  for  expenses?         Ans.  $1050. 

5.  Bought  320  yards  of  calico  @  15  cents,  and  sold  it  at  a 
reduction  of  2i  %  ;  what  was  the  entire  loss? 

6.  A  dealer  having  bought  30  barrels  of  apples  at  $3.50  per 
barrel,  and  shipped  them  at  an  expense  of  $5.38,  to  be  sold  on 
commission  of  5  %,  what  will  be  his  w^hole  loss  if  the  selling 
price  is  10  %  below  the  purchase  price  ?  Ans.  $20.60^. 

7.  Bought  corn  at  $.50  a  bushel;  at  what  price  must  it  be  sold 
to  gain  33J  per  cent.  ? 

8.  Bought  fish  at  $4.25  per  quintal,  and  sold  the  $^me  at  $4.93 ; 
what  was  my  gain  per  cent.  ?  Ans.  16  %. 

9.  Bought  a  hogshead  of  sugar  containing  9  cwt.  44  lb.,  for  $59 ; 
paid  $4.72  for  freight  and  cartage;  at  what  price  per  pound  must 
it  be  sold  to  gain  20  per  cent,  on  the  buying  price  ? 

10.  A  wine  merchant  bought  a  hogshead  of  wine  for  $157.50; 
a  part  having  leaked  out  he  sold  the  remainder  for  $3.32  J  a  gallon, 
and  found  his  loss  to  be  5  per  cent,  on  the  cost ;  how  many  gallons 
leaked  out?  Aois.  18. 

11.  Sold  a  farm  of  106  A.  3  R.  30  P.  for  $9«  an  acre,  and 
gained  18  per  cent,  on  the  cost;  how  much  did  the  whole  farm 
cost?  Ans.  $8700. 

12.  A  lumberman  sold  36840  feet  of  lumber  at  $21.12  per  M, 
and  gained  28  per  cent. ;  how  much  would  he  have  gained  or  lost, 
had  he  sold  it  at  $17  per  M  ?  Ans.  $18.42,  gained. 

13.  A  speculator  bought  shares  in  a  mining  company  when  the 
stock  was  4  %  below  par,  and  sold  the  same  when  it  was  28  % 
below  par;  what  per  cent,  did  he  lose  on  his  investment? 

14.  A  machinist  sold  a  fire  engine  for  $7050,  and  lost  6  per 
cent,  on  its  cost;  for  how  much  ought  he  to  have  sold  it  to  gain 
12}  percent.?  Ans.  $8437.50. 

•  15.  Sold  my  carriage  at  30  per  cent,  gain,  and  with  the  money 
bought  another,  which  I  sold  for;$182,  and  lost  12|  per  cent.; 
how  much  did  each  carriage  coet  me  ?        .        (  Firt5t,  $160  ; 

t  Se«€>nd,  $208. 


PROFIT  AND  LOSS.  289 

*  16.  GafTney,  Burke  &  Co.  bought  a  quantity  of  dry  goods  for 
$6840;  they  sold  }  of  them  at  15  per  cent,  profit,  i  at  18i  per 
cent.,  i  at  20  per  cent.,  and  the  remainder  at  33 J  per  cent,  profit; 
how  much  was-  the  average  gain  per  cent.,  and  how  much  the  whole 
gain?  Ans.   21|     %   gain;  $1482,  entire  gain. 

•17.  If  I  buy  a  piece  of,  land,  and  it  increases  in  value  each 
year  at  the  rate  of  50  per  cent,  on  the  value  of  the  previous  year, 
for  4  years,  and  then  is  worth  $12000,  how  much  did  it  cost  ? 

.  18.  A  Western  merchant  bought  wheat  as  follows  :  600  bushels 
of  red  Southern  @  $1.80,  1200  bushels  of  white  Michigan  @ 
$1.62^,  and  200  bushels  of  Chicago  spring,  @  $1.25.  lie 
shipped  the  whole  to  his  correspondent  in  Buffalo,  who  sold  the 
first  two  kinds  at  an  advance  of  20  %  in  the  price,  and  the  bal- 
ance at  $1.20  per  bushel,  and  deducting  from  the  gross  avails  his 
commission  at  5  %,  and  $254.60  for  expenses,  returned  to  the 
consignor  the  net  proceeds.  What  was  the  rate  of  the  merchant's 
gain  ?  Ans.  ^\  ^c 

19.  A  broker  buys  stock  when  it  is  20  *y^  below  par,  and  sells 
it  when  it  is  16  %  below  par;  w^hat  is  his  rate  of  gain  ? 

20.  A  man  has  5  per  cent,  stock  the  market  value  of  which  is 
78  %  ;  if  he  sells  it,  and  takes  in  exchange  6  %  stock  at  4  ^ 
premium,  what  per  cent,  of  his  annual  income  does  he  lose? 

^  •  21.  A  machinist  sold  24  grain-drills  for  $125  each.  On  one 
half  of  them  he  gained  25  per  cent.,  and  on  the  remainder  he 
lost  25  per  cent.;  did  he  gain  or  lose  on  the  whole,  and  how 
much?  ^ns.  Lost  $200. 

B  22.  Bought  land  at  $30  an  acre ;  how  much  must  1  ask  an  acre, 
that  I  may  abate  25  per  cent,  from  my  asking  price,  and  still  make 
20  per  cent,  on  the  purchase  money  ?  Am.  $48. 

H^  23.  A  salesman  asked  an  advance  of  20  per  cent,  on  the  cost 
of  some  goods,  but  was  obliged  to  sell  at  20  per  cent,  less  than 
his  asking  price;  did  he  gain  or  lose,  and  how  much  per  cent.? 
\^  24.  A  Southern  merchant  ships  to  his  agent  in  IJdston,  a  quan- 
tity  of  sugar  consisting  of  200  bbl.  of  New  Orleans,  each  containing 
216  lb.,  purchased  at  5  cents  per  pound,  and  560  bbl.  of  West 
India,  each  containing  200  lb.,  purchased  at  5 J  cents  per  pound.- 
25  T 


290  PERCENTAGE. 

The  agent's  account  of  sales  shows  a  loss  of  1  %  on  the  New  Or- 
leans, and  a  profit  of  |§  %  on  the  West  India  sugar ;  does  the 
merchant  gain  or  lose  on  the  whole  consignment,  and  what  per 
cent.?  Ans.  Gains  |  %. 

25.  ^A  grocer  sold  a  hogshead  of  molasses  for  $31.50,  which 
was  a  reduction  of  30  %  from  the  prime  cost ;  what  was  the  pur- 
chase price  paid  per  gallon  ? 

26.  A  speculator  sold  stock  at  a  discount  of  7|  %,  and  made 
a  profit  of  5  %  ;  at  what  rate  of  discount  had  he  purchased  the 
stock?  Ans,  12  %. 

27.  A  dry-goods  merchant  sells  delaines  for  2^  cents  per  yard 
more  than  they  cost,  and  realizes  a  profit  of  8  %  ;  what  was  the 
cost  per  yard?  Ans.   $.31|. 

28.  If  I  make  a  profit  of  18|  %  by  selling  broadcloth  for  $.75 
per  yard  above  cost,  how  much  must  I  advance  on  this  price  to 
realize  a  profit  of  31|  %  ? 

29.  A  speculator  gained  SO  %  on  |  of  his  investment,  and  lost 
6  %  on  the  remainder,  and  his  net  profits  were  $720.  What 
would  have  been  his  profits,  had  he  gained  30  ^  on  |  and  lost 
6  %  on  the  remainder?  Ans.  $405. 

30.  A  man  wishing  to  sell  his  real  estate  asked  36  per  cent, 
more  than  it  cost  him,  but  he  finally  sold  it  for  16  per  cent,  less 
than  his  asking  price.  He  gained  by  the  transaction  $740.48. 
How  much  did  the  estate  cost  him,  what  was  his  asking  price,  and 
for  how  much  did  he  sell  it  ? 

Ans,  Cost,  $5200 ;  asking  price,  $7072;  sold  for  $5940.48. 

31.  Sold  I  of  a  barrel  of  beef  for  what  the  whole  barrel  cost; 
what  per  cent,  did  I  gain  on  the  part  sold  ? 

32.  Bought  4  hogsheads  of  molasses,  each  containing  84  gal- 
lons, at  $.37  J  a  gallon,  and  paid  $7.50  for  freight  and  cartage. 
Allowing  5  per  cent,  for  leakage  and  waste,  4  per  cent,  of  the  sales 
for  bad  debts,  and  1  per  cent,  of  the  remainder  for  collecting,  for 
how  much  per  gallon  must  I  sell  it  to  make  a  net  gain  of  25  per 
cent,  on  the  whole  cost?  Ans.  $.55  +  . 


INSURANCE.  291 

INSURANCE. 

4©C5.  Insurance  is  security  guaranteed  by  one  party  to  ano- 
ther, against  loss,  damage,  or  risk.  It  is  of  two  kinds;  insurance 
on  property,  and  insurance  on  life. 

40?^.  The  Insurer  or  Underwriter  is  the  party  taking  the 
risk. 

498.    The  Insured  or  Assured  is  the  party  protected. 

400.    The  Policy  is  the  written  contract  between  the  parties. 

^00.  Premium  is  the  sum  paid  for  insurance.  It  is  always  a 
certain  per  cent,  of  the  sum  insured,  varying  according  to  the 
degree  or  nature  of  risk  assumed,  and  payable  annually  or  at  stated 
intervals. 

NoTKS. — 1.  Insurance  business  is  generally  conducted  by  joint  stock  compa- 
nies, though  Siunetimes  by  individuals. 

2.  A  3/ntnal  luHurauce  company  is  one  in  which  each  person  insured  is  enti- 
tled to  a  share  in  the  profits  of  the  concern. 

3.  The  act  of  insuring  is  sometimes  called  taking  a  risk. 

FIRE   AND    MARINE   INSURANCE.  '" 

^01.  Insurance  on  property  is  of  two  kinds;  Fire  Insurance 
and  Marine  Insurance. 

Fire  Insurance  is  security  against  loss  of  property  by  fire. 

Marine  Insurance  is  security  against  the  loss  of  vessel  or  cavgo 
by  the  casualties  of  navigation. 

^O^.  The  Sum  Covered  by  insurance  is  the  difference  be- 
tween the  sum  insured  and  the  premium  paid. 

Notes. — 1.  As  security  against  fraud,  most  insurance  companies  take  risks  at 
not  more  than  two- thirds  of  the  full  value  of  the  property  insured. 

2.  When  insured  property  suffers  damage  less  than  the  amount  of  the  policy, 
the  insurers  are  required  to  pay  only  the  estimated  loss. 

^03,  The  calculations  in  insurance  are  based  upon  tho  fol- 
lowing relations  : 

I.  Premium  is  perrenta//e  (44:3). 

II.  The  sum  insured  is  the  base  of  premium. 

III.  The  sum  covered  by  insurance  is  difference. 

EXAMPLES    FOR   PRACTICE. 

1 .  What  premium  must  be  paid  for  insuring  mj  stock  of  goods 
to  the  amount  of  ?5760  at  li  %  ? 


292  PERCENTAGE. 

OPERATION.  Analysis.     According  to 

$5760  X  .0125  ==  $72,  Ans.  Prob.  I,  (449),  we  multiply 

$5760,  the  base  of  premium, 
by  .0125,  the  rate,  and  obtain  $72,  the  premium. 

2.  For  what  sum  must  a  granary  be  insured*  at  2  %  in  order  to 
cover  the  loss  of  the  wheat,  valued  at  $1617  ? 

OPERATION.  Analysis.     According  to 

1.00 .02  =  .98  P^^ob.  V,  (453),  we  divide  the 

$1617  ~-  .98  =  $1650,  Ans,  sum  to   be  covered,   $1617, 

which  is  difference,  by  1 
minus  the  rate  of  premium,  and  obtain  $1650,  the  base  of  premium, 
or  the  sum  to  be  insured. 

Proof.  $1650  X  .02  =  $33,  premium ;  $1650  —  $33  =  $1617,  the 
sum  covered. 

3.  What  must  be  paid  for  an  insurance  of  $5860  at  1 J  %  ? 

4.  What  is  the  premium  of  $860  at  }  %  ?  Am.  $4.30. 

5.  What  is  the  premium  for  an  insurance  of  $3500  on  my  house 
and  barn,  at  IJ  %  ?  Ans.  $43.75. 

6.  A  fishing  craft,  insured  for  $10000  at  2J  %,  was  totally 
wrecked ;  how  much  of  the  loss  was  covered  ?         A^is.  $9775. 

7.  A  hotel  valued  at  $10000  has  been  insured  for  $6000  at 
li  %,  $5.50  being  charged  for  the  policy  and  the  survey  of  the 
premises ;  if  it  should  be  destroyed  by  fire,  what  loss  would  the 
owner  suffer?  Ans.  $4080.50. 

8.  A  merchant  whose  stock  in  trade  is  worth  $12000,  gets  the 
goods  insured  for  |  of  their  value,  at  J  %  ;  if  in  a  conflagration 
he  saves  only  $2000  of  the  stock,  what  actual  loss  will  he  sustain  ? 

9.  If  I  take  a  risk  of  $36000  at  2J  %,  and  re-insure  J  of  it 
at  3  %,  what  is  my  balance  of  the  premium?  Ans.  $360. 

10.  I  pay  $12  for  an  insurance  of  $800 ;  what  is  the  rate  of 
premium?  A71S.  \\  ^q. 

11.  A  trader  got  a  shipment  of  500  barrels  of  flour  insured  for 
80  ^c  of  its  cost,  at  3}  %,  paying  $107.25  premium;  at  what 
price  per  barrel  did  he  purchase  the  flour?  Ans.  $8.25. 

12.  The  Astor  Insurance  Company  took  a  risk  of  $16000,  for 
a  premium  of  $280  ;  what  was  the  rate  of  insurance  ? 

13.  A  whaling  merchant  gets  his  vessel  insured  for  $20000  in 


INSURANCE.  293 

the  Gallatin  Company,  at  f  %,  and  for  $30000  in  the  Howard 
Company,  at  J  %  ;  what  rate  of  premium  does  he  pay  on  the  whole 
insurance?  Ans.  |  %. 

14.  If  it  cost  $46.75  to  insure  a  store  for  J  of  its  value,  at  1| 
^,  what  is  the  store  worth?  Ans.  $6800. 

15.  For  what  sum  must  I  get  my  library  insured  at  H  %,  to 
cover  a  loss  of  $7910  ?  Ans.  $8000. 

16.  What  will  be  the  premium  for  insuring  at  2|  %,  to  cover 
$27320?  Ans.  $680. 

17.  A  shipment  of  pork  was  insured  at  4f  %,  to  cover  |  of  its 
value.  The  premium  paid  was  $122.50;  what  was  the  pork 
worth?  Ans.  $4480. 

18.  A  gentleman  obtained  an  insurance  on  his  house  for  f  of 
its  value,  at  li  %  annually.  After  paying  5  instalments  of  pre- 
mium, the  house  was  destroyed  by  fire,  in  congcqueiice  of  which 
he  suffered  a  loss  of  $2940 ;  what  was  the  value  of  the  lum^e  ? 

Ans.  $960.0. 

19.  A  man's  property  is  insured  at  2 J  %  payable  annually;  in 
how  many  years  will  the  premium  equal  the  policy  ? 

20.  A  company  took  a  risk  at  2}  %,  and  re-insured  |  of  it  in 
another  company  at  2  J  %.  The  premium  received  exceeded  the 
premium  paid  by  $72.     What  was  the  amount  of  the  risk? 

21.  The  Commercial  Insurance  Company  issued  a  policy  of 
insurance  on  an  East  India  merchantman  for  f  of  the  estimated 
value  of  ship  and  cargo,  at  4}  %  ,  and  immediately  re-insured  J 
of  the  risk  in  the  Manhattan  Company,  at  3  %.  During  the  out- 
ward voyage  the  ship  was  wrecked,  and  the  Manhattan  Company 
lost  $1350  more  than  the  Commercial  Company;  what  did  the 
owners  lose?  '  Ans.  $40590. 

LIFE   INSURANCE. 

^©-4.   Life  Insurance  is  a  contract  in  which  a  company  agrees 

to  pay  a  certain  sum  of  money  on  the  death  of  an  individual,  in 

consideration  of  an  immediate  payment,  or  of  an  annual  premium 

paid  for  a  term  of  years^  or  during  the  life  of  the  insured.     The 

25* 


294  PERCENTAGE. 

policy  may  be  made  payable  to  the  heirs  of  the  insured,  or  assured, 
person,  or  to  any  one  whom  he  may  designate. 

^^e>.  The  policies  issued  by  life  insurance  companies  are  of 
various  kinds,  the  principal  of  which  are  as  follows : 

1st.  Term  policies,  payable  on  the  death  of  the  insured,  if  the 
death  occurs  during  a  specified  term  of  years ;  these  require  the 
payment  of  an  annual  premium  till  the  policy  matures  or  expires. 

2d.  Life  policies,  payable  on  the  death  of  the  insured,  the  annual 
premium  to  continue  during  life. 

8d.  Life  policies,  payable  on  the  death  of  the  insured,  the  annual 
premium  to  cease  at  a  given  age. 

4th.  Endowment  assurance  policies,  payable  to  the  assured 
person  on  his  attaining  a  given  age,  or  to  his  heirs  if  his  death 
occurs  before  that  age,  annual  premium  being  required  till  the 
policy  matures. 

NoTK.  —  The  premium  on  tlie  firpt  nrd  peconri  cln?ses  of  policies  innj  he  dis- 
cliar<re(l  by  n.  single  payment,  in.•^tea(l  of  annual  installments. 

HOQ.  The  Expectaticn  cf  Life  is  the  average  number  of  years 
of  life  that  remains  to  persons  of  a  given  age,  as  determined  by 
tables  of  mortality. 

^O'^.  The  rates  of  life  insurance,  as  fixed  by  different  com- 
panies, are  based  upon  the  expectation  of  life  and  the  probable 
rates  of  interest  which  money  will  bear  in  future  time. 

•508.  The  rates  of  annual  premium  for  the  assurance  of  $100 
on  a  f^ingle  life,  according  to  the  two  kinds  of  life  jwUcies  (2  and 
8),  as  issued  by  the  Mutual  Life  Insurance  Company  of  New 
York,  are  given  in  the  Life  Table  on  page  291. 

•^01^.  The  rates  of  annual  premium  of  an  assurance  of  $100 
in  the  same  company,  payable  to  the  party  assured  on  his  attain- 
ing the  age  of  40,  45,  50,  55,  60,  or  G5,  or  to  his  representatives, 
in  case  of  death  before  attaining  these  ages  respectively,  are  shown 
in  the  Endowment  Assurance  Table  on  page  292. 

ICoTES.  —  1.  The  tables  of  the  Mutual  Life  Insurance  Companj''  of  New  York 
have  l»een  fele<te<l,  as  furnishing  go(.<l  exjimples  of  a  variety  of  policies;  the 
computationt:  by  any  other  tables  would  not  diliVr  in  any  material  retpect  fn-m 
thot^e  introduced  under  these  tables. 

2.  Since  a  payment  is  made  at  the  issue  of  the  policy,  and  another  at  the  ex- 
piration of  the  first  year,  the  number  of  payments  on  a  policy  will  always  be  1 
more  than  the  number  of  years. 


INSURANCE. 
LIFE    TABLE. 


295 


ANNUAL    PREMIUM    ON   A    POLICY    OF    $100. 


Affeat 

Psvmen's 

Payments 

Payments 

Payments 

Age  at 

issue. 

during  life. 

To  cease  at  65. 

To  cease  at  60. 

To  cease  at  50. 

issue. 

14 

$1.4707 

$1.4999 

$1.5238 

$1.6150 

14 

15 

1.5105 

1.5422 

1.5683 

l.CGSl 

15 

16 

1.5516 

1.5861 

1.C145 

1.7240 

16 

17 

1.5940 

1.C316 

1.6C25 

1.7826 

17 

18 

1.C377 

1.67S6 

1.7124 

1.8444 

18 

19 

l.GS'JO 

1.7275 

1.7644 

1.9096 

19 

20 

1.7296 

1.7782 

1.8186 

1.9785 

20 

21 

1.7780 

1.8310 

1.8753 

2.0516 

21 

22 

1.8280 

l.SSoO 

19344 

2.1292 

22 

23 

1.8798 

1.9131 

1.99C3 

2.2118 

23 

24 

1.9335 

2  0027 

2.0612 

2.3000 

24 

25 

1.9891 

2.0C48 

2.1291 

2.3944 

25 

26 

2.0470 

2.1300 

2.2007 

2.4959 

26 

27 

2.1071 

2.1981 

2.27C1 

'    2.6054 

27 

28 

2.1C96 

2.2095 

2.3555 

2.7238 

28 

29 

2,2346 

2.344i 

2.4305 

2  8525 

29 

30 

23023 

2.4230 

2.5284 

2.9028 

30 

31 

2.3728 

2.5058 

2.6228 

3.14C6 

31 

32 

2.44C4 

2.59.30 

2.7223 

3.31C3 

32 

33 

2.52.32 

2.CS51 

2.8296 

3.5044 

33 

34 

2.C034 

2.7824 

2.9436 

3.7142 

34 

35 

2.6873 

2.8856 

3.0657 

3.9503 

35 

36 

2.7752 

2.9951 

3.1971 

4.2182 

36 

37 

2.8674 

3.1117 

3.5387 

4.5251     - 

37 

38 

2  9641 

3.2361 

3.4919 

4.8807 

38 

39 

3.0C58 

3.3692 

3.6584 

5.2981 

39 

40 

3.1729 

3.5120 

3.8402 

5.7959 

40 

41 

3.2856 

3.GG54 

4.0333 

41 

42 

3.4046 

3.8311 

4.25S8 

42 

43 

3.5303 

4.0106 

4.5021 

43 

44 

3.6632 

4.2055 

4.7735 

44 

45 

3.8038 

4.4181 

5.0782 

45 

46 

3.9530 

4.C512 

5.4235 

46 

47 

4.1111 

4.9075 

5.81  SO 

47 

48 

4.2782 

6.1902 

6.2726 

48 

49 

4.4549 

5.5038 

6.SC32 

49 

50 

4  6417 

5.8536 

7.4317 

50 

51 

4.8393 

6.2470 

51 

52 

5.0486 

6.6935 

52 

53 

6.2708 

7.2oei 

53 

54 

5.50C7 

7.8017 

54 

£5 

6.7577 

8.5048 

55 

296 


PERCENTAGE. 
ENDOWMENT   ASSURANCE   TABLE. 


ANNUAL  PREMIUM   ON   A   FOLIC! 

OF    $100. 

Age  at 

Policy  due 

Policy  due 

Policy  due 

Policy  due 

Policy  due 

Policy  due 

Age  at  ' 

issue. 

at  40. 

at  45. 

at  50. 

at  55. 

at  60. 

at  65. 

issue. 

14 

$2,475 

$2,113 

$1,868 

$1,704 

14 

15 

2.587 

2.197 

1.935 

1.759 

15 

16 

$3,356 

2.707 

2.2S5 

2.C04 

1.816 

$1,694 

16 

17 

3.545 

2.835 

2.379 

2.077 

1.876 

1.746 

17 

IS 

3.752 

2.937 

2.478 

2.153 

1.939 

1.799 

18 

19 

3.978 

3.122 

2.585 

2.234 

2.004 

1.855 

19 

20 

4.228 

3.283 

2.698 

2.320 

2.073 

1.914 

20 

21 

4.504 

3.458 

2.819 

2.410 

2.145 

1.974 

21 

22 

4.812 

3.648 

2.919 

2.506 

2.220 

2.03S 

22 

23 

5156 

3855 

3.089 

2.608 

2.300 

2.104 

23 

24 

6.544 

4.0S3 

3.239 

2.717 

2.384 

2.174 

24 

25 

5.985 

4.333 

3.402 

2.832 

2.473 

2.247 

25 

26 

6.489 

4.611 

3.578 

2.956 

2.567 

2.323 

26 

27 

7.082 

4.920 

3.770 

3.088 

2.666 

2.404 

27 

2S 

7.752 

5.265 

3.979 

3.231 

2.772 

2.489 

28 

29 

8.558 

5.654 

4.208 

3.384 

2.884 

2.578 

29 

SO 

9.526 

6.096 

4.4G1 

3.549 

3.004 

2.672 

30 

31 

6.G01 

4.740 

3.728 

3.132 

2.772 

31 

32 

7.185 

5.051 

3.923 

3.268 

2.877 

32 

33 

7.866 

5.398 

4.135 

3.415 

2.989 

33 

34 

8.673 

5.789 

4.308 

3.573 

3.108 

34 

35 

9.641 

6.232 

4.624 

3.743 

3.235 

35 

36 

6.739 

4.900 

3.928 

3.370 

36 

37 

7.325 

6.221 

4.128 

3.515 

37 

38 

8.008 

5.572 

4.317 

3.670 

38 

SO 

8.816 

5.967 

4.5S6 

3.837 

39 

40 

9.787 

6.415 

4.849 

4.017 

40 

41 

C.9J7 

5,139 

4.212 

41 

42 

7.518 

,6,462 

4.424 

42 

43 

8  207 

5.822 

4.G55 

43 

44 

9.022 

6.227 

4.908 

44 

45 

10.000 

6.CS6 

5.185 

45 

46 

7.210 

5.491 

46 

47 

7.813 

5.S30 

47 

48 

8.515 

e.-jos 

48 

49 

9.343 

6.630 

49 

50 

10.332 

7.105 

50 

51 

11.530 

7.045. 

61 

52 

8.':65 

62 

53 

8.983 

53 

54 

9.826 

54 

55 

10.S31 

55 

INSURANCE.  297 

EXAMPLES    FOR   PRACTICE. 

1.  What  sum  must  a  man  fay  annually  to  the  Mutual  Insurance 
of  New  York,  for  a  life  policy  of  825C0,  his  age  being  33  years 
at  the  issue  of  the  policy  ? 

oPERATiox.  Analysis.    Wo  multl- 

82500  X  .025232  =  §63.08,  Ans.       ply  the  face  of  the  policy, 

$2500,   by  the  rate  per 

cent,  found  opposite  33  years  in  the  Life  Table,  expressed  decimally, 

and  obtain  §03.08,  the  annual  premium  required. 

NoTK.  —  The  examples  which  follow  all  refer  to  the  rates  given  in  the  pre- 
ceding tables. 

2.  A  man  at  30  years  of  age  takes  a  policy  for  ^2000,  the  pay- 
ments of  premium  to  cease  at  50 ;  if  he  survives  that  age,  how 
much  more  money  will  he  receive  from  the  company  than  he  pays 
to  the  company  ?  Ans.   $743,024. 

3.  What  annual  premium  must  a  man  pay  during  life,  com- 
mencing at  the  age  of  50,  to  secure  S3000  at  his  death '/ 

4.  A  gentleman  at  the  age  of  3G  gets  his  life  insured  for  §1500, 
premium  to  cease  at  the  age  of  60  ;  if  he  dies  at  52,  how  much 
more  will  his  family  receive  than  has  been  paid  out  in  premiums  ? 

5.  A  clergyman  wishing  to  secure  an  income  to  his  family  after 
his  death,  had  his  life  insured  at  the  age  of  54,  in  the  sum  of 
§3500,  premium  payable  during  life;  his  decease  took  place  at 
the  age  of  72.  How  much  more  would  have  been  saved  to  his 
family  if  he  had  taken,  instead,  a  policy  for  the  same  amount,  v^'ith 
payments  of  premium  to  cease  at  65?  Ans.  §3S5.24. 

6.  IIow  much  more  premium  will  be  required  to  secure  an  en- 
dowment of  §1200  at  40,  by  taking  out  a  policy  at  the  age  of  30, 
than  if  the  policy  be  taken  at  24  ?  Ans.  §126.456. 

7.  A  man  37  years  old  took  an  endowment  assurance  policy 
for  §750,  due  at  the  age  of  50,  and  died  when  49  years  old ;  how 
much  more  would  his  heirs  have  realized  if  he  had  taken  a  life 
policy  for  the  same  amount,  with  payment  to  cease  at  50  ? 

8.  A  has  his  life  insured  at  the  age  of  20,  and  B  has  his  in- 
sured at  the  age  of  30,  each  taking  a  life  policy  requiring  annual 
payments  of  premium  during  life ;  what  will  be  the  age  of  each 


298  PERCENTAGE. 

when  the  amount  of  premiunx  paid  shall  exceed  the  face  of  his 
policy?  Ans.  A,  77  years;  B,  73  years. 

9.  What  is  the  whole  amount  of  premiums  that  must  be  paid 
to  secure  an  endowment  of  §1000  at  the  age  of  GO,  the  policy 
being  issued  at  the  age  of  45  ?  Ans.  $1069.70. 

10.  A  person  at  the  age  of  34  had  his  life  insured  in  the  sum 
of  $600,  the  premium  to  cease  at  50.  When  he  died,  there  was 
a  net  gain  to  his  family  of  §421.72 ;  how  many  payments  of  pre- 
mium had  he  made  ?  Ans.  8. 

11.  A  gentleman  obtained  an  insurance  on  his  life  at  the  age 
of  29,  and  died  at  the  age  of  40 ;  the  policy  taken  required  an- 
nual payments  of  premium  during  life,  and  secured  to  his  heirs 
$1829.62  more  than  the  whole  premium  paid.  Required  the  face 
of  the  policy.  Ans.  $2500. 

TAXES. 

510*  A  Tax  is  a  sum  of  money  assessed  on  the  person  or  pro- 
perty of  an  individual,  for  public  purposes. 

^11,  A  Poll  Tax  is  a  certain  sum  required  of  each  male  citi- 
zen liable  to  taxation,  without  regard  to  his  property.  Each  person 
so  taxed  is  called  2i  poll. 

511^.  A  Property  Tax  is  a  sum  required  of  each  person  own- 
ing property,  and  is  always  a  certain  ^^er  cent,  of  the  estimated 
value  of  his  property. 

•IIS,  An  Assessment  Eoll  is  a  list  or  schedule  containing  the 
names  of  all  the  persons  liable  to  taxation  in  the  district  or  com- 
pany to  be  assessed,  and  the  valuation  of  each  person's  taxable 
property. 

qSIJ:.  Assessors  are  the  persons  appointed  to  prepare  the  as- 
sessment roll,  and  apportion  the  taxes. 

1.  In  a  certain  town  a  tax  of  $4000  is  to  be  assessed.  There 
are  400  polls  to  be  assessed  8.50  each,  and  the  valuation  of  the 
taxable  property,  as  shown  by  the  assessment  roll,  is  §950000 ; 
what  will  be  the  property  tax  on  $1,  and  how  much  will  be  A's 
tax,  whose  property  is  valued  *at  $3500,  and  who  pays  for  3  polls? 


TAXES. 


299 


OPERATION. 

$    .50  X     400^=  $200,  amount  assessed  on  the  polls. 
84000  —  8200  =  ^3800,  amount  to  be  assessed  on  property. 
§3800  -f-  §950000  =  .004,  rate  oftaxutiun  ; 
§3500  X  -004  =  §14,        A's  property  tax; 
§    .50  X  3    =         L50,  A's  poll  tax; 

§15.50,  amount  of  A's  tax.     Hence  the 
Rule.     I.   Find  the  amount  of  jxM  tax,  if  any,  and  suhtract  it 
from  the  whole  fax  to  he  assessed ;  the  remainder  will  he  the  proj)- 
erty  tax. 

II.  Divide  the  property  tax  hy  the  whole  amount  of  taxahle 
property  ;  the  quotient  will  he  the  rate  of  taxation. 

III.  Multiply  eaeh  man^s  taxahle  property  hy  the  rate  of  taxa- 
tion, and  to  the  produet  add  his  poll  tax,  if  any  ;  the  result  will  he 
the  whole  amount  of  his  tax. 

Note. — When  n.  tax  is  to  he  apportioned  among  a  large  number  of  individuals, 
the  operation  is  greatly  facilitated  by  first  finding  the  tax  on  $1,  $2,  $3,  etc.,  to 
$9;  then  on  $10,  $20,  $30,  etc.,  to  $90,  and  so  on,  and  arranging  the  results  as 
in  the  following 

TABLE. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1 

$.004 

$10 

$.04 

$100 

$  .40 

$1000 

$4.00 

2 

.008 

20 

.08 

200 

.80 

2000 

8. 

3 

.012 

30 

.12 

300 

1.20 

3000 

12. 

4 

.016 

40 

.16 

400 

1.60 

4000 

16. 

5 

.020 

50 

.20 

500 

2.00 

5000 

29. 

6 

.024 

60 

.24 

600 

2.40 

6000 

24. 

r 

.028 

70 

.28 

700 

2.80 

i  7000 

28. 

8 

.0.']2 

80 

.32 

800 

3.20 

8000 

32. 

9 

.036 

90 

.36 

900 

3.60 

9000 

36. 

EXAMPLES    FOR    PRACTICE. 


1.  According  to  the  conditions  of  the  last  exanjple,  what  would 
be  the  tax  of  a  person  whose  property  was  valued  at  24G5,  and 
who  pays  for  2  polls 't 


)0                                            PERCENTAGE. 

OPERATION. 

From  the  table  we  find  that 

The  tax  on  $2000 

is 

$8.00 

u       a      a         400 

a 

1.60 

ic        ic      a            60 

iC 

.24 

((       a      a              5 

u 

.02 

And    "       "      "             2 

polls 

u 

1.00 

Whole  tax  "  §10.86,  Ans. 

2.  What  would  A's  tax  be,  who  is  assessed  for  S8530,  and  3 
polls?  Ans.  835.62. 

3.  How  much  will  C's  tax  be,  who  is  assessed  for  $987,  and  1 
poll?  Ans,  $4,448. 

4.  The  estimated  expenses  of  a  certain  town  for  one  year  are 
§6319,  and  the  balance  on  hand  in  the  public  treasury  is  $354. 
There  are  2156  polls  to  be  assessed  at  $.25  each,  and  taxable  pro- 
perty to  the  amount  of  $1864000.  Besides  the  town  tax,  there 
is  a  county  tax  of  IJ  mills  on  a  dollar,  and  a  State  tax  of  i  of  a 
mill  on  a  dollar.  Required  the  whole  amount  of  A's  tax,  whose 
property  is  valued  at  $32560,  and  who  pays  for  3  polls. 

5.  Wh-at  does  a  non-resident  pay,  who  owns  property  in  the 
same  town  to  the  amount  of  $16840  ?  Ans.  $79.99. 

6.  What  sum  must  be  assessed  in  order  to  raise  a  net  amount 
of  $5561.50,  and  pay  the  commission  for  collecting  at  2  %. 

Note.  —  Since  the  base  of  the  collector's  commission  is  the  sum  collected, 
(446),  the  question  is  an  example  under  Problem  V  of  Percentage. 

7.  In  a  certain  district  a  school  house  is  to  be  built  at  an  ex- 
pense of  $9120,  to  be  defrayed  by  a  tax  upon  property  valued  at 
§1536000.  What  shall  be  the  rate  of  taxation  to  cover  both  the 
cost  of  the  school  house,  and  the  collector's  commission  at  5  ^  ? 

8.  The  expenses  of  a  school  for  one  term  were  $1200  for 
salary  of  teachers,  $57.65  for  fuel,  and  $38.25  for  incidentals ;  the 
money  received  from  the  school  fund  was  $257.75,  and  the  remain- 
ing part  of  the  expense  was  paid  by  a  rate-bill.  If  the  aggregate 
attendance  was  9568  days,  what  was  A's  tax,  who  sent  4  pupils  46 
days  each?  Ans.  $19.96+. 

9.  The  expense   of  building  a  public   bridge,  vras   $1260.52 


GENERAL  AVERAGE.  801 

which  was  defrayed  by  a  tax  upon  the  property  of  the  town.  The 
rate  of  taxation  was  3}  mills  on  one  dollar,  and  the  collector's 
commission  was  3J  5^  ;  what  was  the  valuation  of  the  property  ? 

Ans,  §401920. 


GENERAL  AVERAGE. 

5t3»  General  Average  is  a  method  of  computing  the  loss  to 
be  sustained  by  the  proprietors  of  the  ship,  freight,  and  cargo, 
respectively,  when,  in  a  case  of  common  peril  at  sea,  any  portion 
of  the  property  has  been  sacrificed  or  damaged  for  the  common 
safety. 

^16,  The  Contributory  Interests  are  the  three  kinds  of  prop- 
erty which  are  taxed  to  cover  the  loss.     These  are, 

1st    The  vessel,  at  its  value  before  the  loss. 

2d.  The  freight,  less  J  as  an  allowance  for  seamen's  wages. 

3d.  The  cargo,  including  the  part  sacrificed,  at  its  market  value 
in  the  port  of  destination. 

Note.  — In  New  York  only  ^  of  the  freight  is  made  contributory  to  the  loss. 

5t7*    JGttson  is  the  portion  of  goods  thrown  overboard. 

t518.    The  loss  which  is  subject  to  general  average  includes, 

1st.  Jettson,  or  property  thrown  overboard. 

2d.  Repairs  to  the  vessel,  less  i  on  account  of  the  superior 
worth  of  the  new  articles  furnished. 

Sd.  Expense  of  detention  to  which  the  vessel  is  subject  in  port. 

1.  The  ship  Nelson,  valued  at  S52000,  and  having  on  board  a 
cargo  worth  ?  18000,  on  which  the  freight  was  $3000,  threw  over- 
board a  portion  of  the  goods  valued  at  $5000,  to  escape  wreck  in 
a  storm;  she  then  put  into  port,  and  underwent  repairs  amounting 
to  $1200,  the  expenses  of  detention  being  $350.  What  portion 
of  the  loss  will  be  sustained  by  each  of  the  three  contributing 
interests  ?  What  will  be  paid  or  received  by  the  owners  of  the 
ship  and  freight  ?  What  by  A,  who  owned  $8000  of  the  cari^o, 
including  $35,00  of  the  portion  sacrificed,  and  by  B,  who  owned 
$6000  of  the  cargo,  including  $1500  of  the  portioi*.  sacrificed,  and 
by  C,  who  owned  $4000,  or  the  residue  of  the  cargo  ? 
26 


302  PERCENTAGE. 

OPERATION. 
LOSSES.  COXTRIBUTORY    INTERESTS. 

Jettson, $5000  Yessel, $52000 

Repairs,  less  i, 800  Freight,  less  J, 2400 

Cost  of  detention, 350  Cargo, 18000 

Total, $G150  Total, $72400 

$0150  ^     $72400  ==  .0849447+,  rate  per  cent,  of  loss. 
$52000  X  .0849447  =  $4417.13,      payable  by  vessel. 
2400  X  .0849447  =      203.87,  "         "    freight. 

18000  X  .0849447  =    1529.00,  "         *'    cargo. 

$0150.00,     Total  contribution. 

$8000  X  .0849447  =  $079.50,  payable  by  A. 
0000  X  .0849447  =  509.07,  ''  ''  B. 
4000  X  .0849447  =    339.78,         "         "  C. 

$4417.13  +  $203.87  =  $4021.00,  payable  by  owners  of  vessel  and  freight. 
800.00+    350.00=-    1150.00,       "      to 
4021.00—1150.00=    3471.00,  balance  pavable  bv  ship  owners. 
3500.00—079.50=2820.44,       "        receivable" by  A. 
1500.00—  509.07=     990.33,       "  ''  '^   B. 

Hence  the  following 

BuLE.  I.  Divide  the  sum  of  the  losses  hy  the  sum  of  the  con- 
trihutory  interests  ;  the  quotient  will  he  the  rate  of  contribution. 

II.  Multiply  each  contributory  interest  by  the  rate;  the  products 
will  he  the  respective  contributions  to  the  loss. 

EXAMPLES    FOR   PRACTICE. 

1.  The  ship  Nevada,  in  distress  at  sea,  cut  away  her  mainmast, 
and  cast  OYerboard  }  of  her  cargo,  and  then  put  into  Havana  to 
refit;  the  repairs  cost  S1500,  and  the  necessary  expenses  of  deten- 
tion were  $420.  The  ship  was  owned  and  sent  to  sea  by  George 
Law,  and  was  valued  at  $25000  ]  the  cargo  was  owned  by  Hayden 
&  Co.,  and  consisted  of  2800  barrels  of  flour,  valued  at  $9  per 
barrel,  upon  which  the  freight  was  $4200.  In  the  adjustment  of 
the  loss  by  general  average,  how  much  was  due  from  Law  to 
Ilayden&Co.?  Ans.  $2629.36. 

2.  A  coasting  vessel  valued  at  $28000,  having  been  disabled  in 
a  storm,  entered  port,  and  was  refitted  at  an  expense  of  $270  for 
repairs,  and  $120   for  board  of  seamen,  pilotage,  and   dockage. 


CUSTOM  HOUSE  BUSINESS.  303 

Of  the  cargo,  valued  at  §5000,  $2400  belonged  to  A,  $1850  to  B, 
and  8750  to  C  ;  and  the  amount  sacrificed  for  the  ship's  safety 
was  $1400  of  A's  property,  and  $170  of  B's;  the  gross  chargos 
for  freight  were  $1500.  Required  the  balance,  payable  or  re- 
ceivable, by  each  of  the  parties,  the  loss  being  apportioned  by 
general  average. 

J       I  $1295  payable  by  ship  owners;  $1268  receivable  by  \; 
""''{   41.25      ^'      "  C;  68.25        "         ''  B. 

CUSTOM  HOUSE  BUSINESS. 

«dl9.  Duties,  or  Customs,  are  taxes  levied  on  imported  goods, 
for  the  support  of  government  and  the  protection  of  home  industry. 

tiSO.  A  Custom  House  is  an  office  established  by  government 
for  the  transaction  of  business  relating  to  duties. 

It  is  lawful  to  introduce  merchandise  into'  a  country  only  at 

points  where  custom  houses    are   established.     A    seaport    town 

having  a  custom   house,  is  called  a  port  of  entry.      To  carry  on 

foreign  commerce  secretly,  without  paying  the  duties  imposed  by 

law,  is  smuggling. 

Note.— Customs  or  duties  form  the  principal  source  of  revenue  to  the  General 
Government  of  the  United  States;  by  increasing  the  price  of  imported  goods 
they  operate  as  an  indirect  tax  upon  consumers,  instead  of  a  general  direct  tax. 

tiSl,    Duties  are  of  two  kinds  —  Ad  Valorem  and  Specific. 

Ad  Valorem  Duty  is  a  sum  computed  on  the  cost  of  the  goods 
in  the  country  from  which  they  were  imported. 

Specific  Duty  is  a  sum  computed  on  the  weight  or  measure  of 
the  goods,  without  regard  to  their  cost. 

^SS«  An  Invoice  is  a  bill  of  goods  imported,  showing  the 
quantity  and  price  of  each  kind. 

t5tlS.  By  the  New  Tariff  Act,  approved  March  2,  1857,  all 
duties  taken  at  the  U.  S.  custom  houses  are  ad  valorem.  The 
principal  articles  of  import  arc  classified,  and  a  fixed  rate  is  im- 
posed upon  each  list  or  schedule,  certain  articles  being  excepted 
and  entered  free. 

In  collecting  customs  it  is  the  design  of  government  to  tax 
only  so  much  of  the  merchandise  as  will  be  available  to  the  im- 


S04  PERCENTAGE. 

porter  in  the  market.  The  goods  are  weighed,  measured,  gauged, 
or  inspected,  in  order  to  ascertain  the  actual  quantity  received  in 
port;  and  an  allowance  is  made  in  every  case  of  waste,  loss,  or 
damage. 

5^-i.  Tare  is  an  allowance  for  the  weight  of  the  box  or  the 
covering  that  contains  the  goods.  It  is  ascertained,  if  necessary, 
by  actually  weighing  one  or  more  of  the  empty  boxes,  casks,  or 
coverings.  In  common  articles  of  importation,  it  is  sometimes 
computed  at  a  certain  per  cent,  previously  ascertained  by  frequent 
trials  by  weighing. 

53^.  Leakage  is  an  allowance  on  liquors  imported  in  casks 
or  barrels,  and  is  ascertained  by  gauging  the  cask  or  barrel  in 
which  the  liquor  is  imported. 

53Q,  Breakage  is  an  allowance  on  liquors  imported  in 
bottles. 

537.  Gross  Weight  or  Value  is  the  weight  or  value  of  the 
goods  before  any  allowance  has  been  made. 

5Q8.  Net  Y/eight  or  Value  is  the  weight  or  value  of  the 
goods  after  all  allowances  have  been  deducted. 

NoTKS.  —  1.  Draft  is  an  allowance  for  the  wa.^te  of  certain  article?,  and  is 
innde  only  for  statiHtt'cal  purpoHea  ;  it  does  not  affect  the  amount  of  duty. 

2.  Long  ton  rnejij^ure  isj  employed  in  the  custom  houses  of  the  United  States, 
in  estimating  goods  by  the  ton  or  hundred  weight. 

The  rates  of  this  allowance  are  as  follows : 

On    112  lb 1  lb. 

Above  112  lb.  and  not  exceeding  224  lb.,  2  lb. 

"   224  lb.  ''     "     "     336  lb.,  3  lb. 

«   336  lb.  *'     "     "    1120  lb.,  4  1b. 

"  1120  lb.  *'  "     "    2016  lb.,  7  lb. 

"  2016  lb 9  lb. 

tlS9«  In  all  calculations  where  ad  valorem  duties  are  consid- 
ered, 

I.  The  net  value  of  the  merchandise  is  the  worth  of  the  net 
weight  or  quantity  at  the  invoice  price,  allowance  being  made  in 
cases  of  damage. 

II.  The  duty  is  computed  at  a  certain  legal  per  cent,  on  the 

net  value  of  the  merchandise. 

Note.  —  In  the  following  examples  the  legal  rates  of  duty,  according  to  the 
New  Tariff  Act,  are  given. 


CUSTOM  HOUSE  BUSINESS.  305 


EXAMPLES    FOR   PRACTICE. 


1.  "What  is  the  duty,  at  24  %,  on  an  invoice  of  cassimere  goods 

which  cost  $750  ? 

Analysis.     According  to  Prob.  I, 

OPERATION.  ^^^gj^  ^^^  multiply  the  invoice,  $750, 

§750  X  .2-4  =  $180  which  is  the  base  of  the  duty,  by  the 

given  rate^  and  obtain  the  duty,  $180. 

2.  The  gross  weight  of  3  hogsheads  of  sugar  is  1024  lb.,  1016 
lb.,  and  1020  lb.  respectively;  t]ie  invoice  price  of  the  sugar  7^ 
cents,  and  the  allowance  for  tare  80  lb.  per  hogshead;  what  is  the 
duty,  at  24  %  ? 

OPERATION.  Analysis.   We  first  find 

1024  the   gross   weight   of   the 

1016  three    hhd.    from   which 

^Q^Q  we  subtract  the  tare,  and 

3060,  gross  weight.  obtain  2820  lb.,   the  net 

80  X  3  =     240,  tare.  weight.    We  next  find  the 

omrv         .        •   1  i^  value  of  the   net  weight, 

2820,  net  weight.  .   ^,         ,      ^,       .    ^.  ' 

(^(x'7\  at  7^   cents,    the    invoice 

price,   and   then   compute 


'^ 


$211.50,  net  value.  the  duty  at  24  %  on  this 

.24  value,  and  obtain  $50.76, 


$50.7600,  duty.  *^^  ^^^3^  required. 

3.  Having  paid  the  duty  at  8  %  on  a  quantity  of  Malaga 
raisins,  I  fi.nd  that  the  whole  cost  in  store,  besides  freight,  is  $378 ; 
what  were  the  raisins  invoiced  at  ? 

Analysis.      According  to   Prob. 
ERATiON.       ^  jy^   (452),  we  divide   the   amount^ 

$378  ^  1.08  =  $350  $378^  by  1  plus  the  rate,  1.08,  and 

obtain  the  basCj  or  invoice,  $350. 

4.  A  Boston  jeweler  orders  from  Lubec  a  quantity  of  watch 
movements,  amounting  to  $2780 ;  what  will  be  the  duty,  at  4  %  ? 

5.  What  will  be  the  duty  at  15  %  on  1200  lb.  of  tapioca,  in- 
voiced at  5^  cents  per  pound  ?  Aiis.  $9.90. 

6.  AVhat  is  the  duty  at  15  %  on  54  boxes  of  candles,  each 
weighing  1  cwt.,  invoiced  at  8|  cents  per  pound,  allowing  tare  at 
3^  per  cent.  ? 

2«* 


806  PERCENTAGE. 

7.  A  merchant  imported  50  casks  of  port  wine,  each  contain- 
ing originally  oG  gallons,  invoiced  at  ^2.50  per  gallon.  He  paid 
freight  at  §1.30  per  cask,  and  duty  at  30  %,  1^  %  leakage  being 
allowed  at  the  custom  house,  and  §8.50  for  cartage ;  what  did  the 
wine  cost  him  in  store  ?  Ans.  §5903.25. 

8.  A  liquor  dealer  receives  an  invoice  of  120  dozen  bottles  of 
porter,  rated  at  §1.25  per  dozen;  if  2  %  of  the  bottles  are  found 
broken,  what  will  be  the  duty  at  24  %  ?  Am.  §35.28. 

9.  The  duty  at  19  %  on  an  importation  of  Denmark  satin  was 
$G19.40;  what  was  the  invoice  of  the  goods?        Aiis.  §3260. 

10.  The  duty  on  600  drums  of  figs,  each  containing  14  lb., 
invoiced  at  5^  cents  per  pound,  was  §35.28;  required,  the  rate 
of  duty.  Ans.  8  %. 

11.  A  merchant  in  New  York  imports  from  Havana  200  hhd. 
of  W.  I.  molasses,  each  containing  63  gallons,  invoiced  at  §.30 
per  gallon;  150  hhd.  of  B.  coffee  sugar,  each  containing  500 
pounds,  invoiced  at  §.05  per  pound ;  80  boxes  of  lemons,  invoiced 
at  §2.50  per  box;  and  75  boxes  of  sweet  oranges,  invojced  at 
§3.00  per  box.  What  was  the  whole  amount  of  duty,  estimated 
at  24  ^c  on  molasses  and  sugar,  and  at  8  %  on  lemons  and  oranges  ? 

Ans.  §1841.20. 

12.  A  merchant  imported  56  casks  of  wine,  each  containing  36 
gallons  net,  the  duty  at  30  %  amounting  to  §907.20  ;  at  what 
price  per  gallon  was  the  wine  invoiced  ? 

13.  The  duty  on  an  invoice  of  French  lace  goods  at  24  %,  was 
§132,  an  allowance  of  12  %  having  been  made  at  the  custom 
house  for  damage  received  since  the  goods  were  shipped ;  what 
was  the  cost  or  invoice  of  the  goods.  Ans.  §625. 

14.  A  quantity  of  Yalencias,  invoiced  at  §1654,  cost  me 
§1980.50  in  store,  after  paying  the  duties  and  §12.24  for  freight; 
what  was  the  rate  of  duty? 

15.  The  duty  on  an  importation  of  Bay  rum,  after  allowing 
2  ^  for  breakage,  was  §823.20,  and  the  invoice  price  of  the  rum 
was  §.25  per  bottle;  how  many  dozen  bottles  did  the  importer 
receive^duty  at  24  %  ?  Ans.  1143-|  doz. 


SIMPLE  INTEREST. 


807 


SIMPLE    INTEREST. 

^30.    Interest  Is  a*sum  paid  for  the  use  of  money. 

^31.  Principal  is  the  sum  for  the  use  of  which  interest  is 
paid. 

*iSQ.  Hate  per  cent,  per  annum  is  the  sum  per  cent,  paid 
for  the  use  of  any  principal  for  one  year, 

KoTK.  —  The  rate  per  cent,  is  commonly  expressed  decimally  as  hundredths 

(442). 

•133,    Amount  Is  the  sum  of  the  principal  and  interest. 

•534:.  Simple  Interest  is  the  sum  paid  for  the  use  of  the 
principal  only,  during  the  whole  time  of  the  loan  or  credit. 

•53cd.  Legal  Interest  is  the  rate  per  cent,  established  by  law. 
It  varies  in  different  States,  as  follows . 

7  per  cent. 


Alabama, 8  per  cent, 

Arkansas, G  '*  " 

California, 10  ''  *' 

Connecticut, G  "  " 

Delaware,.. G  "  *' 

Dist.  of  Columbia,  . .  G  *'  *' 

Florida, 8  ''  '' 

Georgia, 7  '*  ** 

Illinois G  "  " 

Indiana, G  **  " 

Iowa, 6  "  ** 

Kentucky G  '*  " 

Louisiana, 5  "  *' 

Maine, G  ''  '' 

Maryland, G  *'  " 

Massachusetts, ......  G  ''  " 

Michigan, 7  ''  '* 


Minnesota,  . . 
Mississippi,  . 

Missouri, .    G 

New  Hampshire, 6 

New  Jersey, 6 

New  York, 7 

North  Carolina, 6 

Ohio, G 

Pennsylvania, G 

Rhode  Island, G 

South  Carolina, 7 

Tennessee, G 

Texas, 8 

United  States  (debts),..  G 

Vermont, G 

Virginia, G 

Wisconsin, 7 


Notes. — 1.  The  lecrnl  rate  in  Canada,  Nova  Scotia,  and  Ireland  is  0  per  cent., 
and  in  England  and  France  5  per  cent. 

2.  When  the  rate  per  cent,  is  not  specified  in  accounts,  notes,  mortgages, 
contracts,  etc.,  the  legal  rate  is  always  understood. 

.3.  In  some  States  the  laws  allow  parties  to  give  nnd  take  higher  rates,  by 
special  agreement. 

4.  Book  accounts  bear  interest  after  the  expiration  of  the  term  of  credit,  and 
notes  are  on  interest  after  they  become  due,  though  no  mention  of  interest  ba 
made  in  them. 

5.  If  notes  arc  to  draw  interest  from  their  date,  or  from  a  given  time  after 
date,  tlie  fact  must  be  so  stated  in  the  body  of  the  notes. 

«I3G«  Usury  is  illegal  interest,  or  a  greater  per  cent,  than  the 
legal  late. 

Note. — The  taking  of  usury  is  prohibited,  under  various  penalties,  in  difFcrcnt 

States. 


808  PERCENTAGE. 

^37*  In  the  operations  of  interest  there  are  five  parts  or  ele- 
ments, namely : 

I.  Rate  per  cent,  per  annum ;  which  is  the  fraction  or  decimal 
denoting  how  many  hundredths  of  a  number  or  sum  of  money  are 
to  be  taken  for  a  period  of  1  year. 

.II.  Interest;  which  is  the  whole  sum  taken  for  the  whole  period 
of  time,  whatever  it  may  be. 

III.  Principal ;  which  is  the  base  or  sum  on  which  interest  is 
computed. 

ly.  Amount;  which  is  the  sum  of  principal  and  interest;  and 

Y.  Time. 

TO    COMPUTE    INTEREST. 
CASE    I. 

^3§,  To  find  tlie  interest  on  any  sum,  at  any  rate 
per  cent,  per  annum,  for  years  and  months. 

Analysis.  In  interest,  any  rate  per  cent,  i^  confined  to  1  year. 
Therefore,  if  the  time  be  more  than  1  year,  the  per  cent,  will  be  greater 
than  the  rate  per  cent,  per  annum,  and  if  the  time  be  less  than  1 
year,  the  per  cent,  will  be  less  than  the  rate  per  cent,  per  annum. 
From  these  facts,  we  deduce  the  following  principles  : 

I.  If  the  rate  per  cent,  per  annum  be  multiplied  by  the  time, 
expressed  in  3'ears  and  fractions  or  decimals  of  a  year,  the  product 
will  be  the  rate  for  the  required  time.     And 

II.  If  the  principal  be  multiplied  by  the  rate  for  the  required 
time,  the  product  will  be  the  required  interest.     Hence 

III.  Interest  is  always  the  product  of  three  factors,  namely, 
rate  per  cent,  per  annum,  time,  and  principal. 

In  computing  interest  the  three  factors  may  be  taken  in  any  order ; 
thus,  if  the  principal  be  multiplied  by  the  rate  per  cent,  per  annum, 
the  product  will  be  the  interest  for  1  year ;  and  if  the  interest  for  1 
year  be  multiplied  bj  the  time  expressed  in  years,  the  result  will  be 
the  required  interest.     Hence  the  following 

HuLE.  I.  Multijyly  the  principal  ly  the  rate  per  cent.y  and  the 
product  Vjill  he  the  interest  for  1  i/ear. 

II.  Multiply  this  product  hy  the  time  in  years  and  fractions  of 
a  year;  the  result  will  he  the  required  interest. 


SIMPLE  INTEREST.  309 

Or,  MultipJij  togetlier  the  rate  per  cent,  jper  anmim^  timcy  and 
principcdj  in  such  order  as  is  most  convenient ;  the  continued  pro- 
duct will  he  the  required  interest. 

CASE   II. 

^39.  To  find  the  interest  on  any  sum,  for  any  time, 
at  any  rate  per  cent. 

The  analysis  of  our  rule  is  based  upon  the  following 

Obvious  Relations  between  Time  and  Interest. 
I    The  interest  on  any  sum  for  1  year  at  1  per  cent.,  is  .01  of 
that  sum,  and  is  equal  to  the  principal  with  the  separatrix  re- 
moved two  places  to  the  left. 

II.  A  month  being  j^^  of  a  year,  -'3^  of  the  interest  on  any  sum 
for  1  year  is  the  interest  for  1  month. 

III.  The  interest  on  any  sum  for  3  days  is  ^\  =  J^  =  ,1  of  the 
interest  for  1  month,  and  any  number  of  days  may  readily  be  re- 
duced to  tenths  of  a  month  by  dividing  by  3. 

IV.  The  interest  on  any  sum  for  1  month,  multiplied  by  any 
given  time  expressed  in  months  and  tenths  of  a  month,  will  pro- 
duce the  required  interest. 

These  principles  are  sufficient  to  establish  the  following 

Rule.  I.  To  find  the  interest  for  1  yr.  at  1  %  : — Remove  the 
separatrix  in  the  given  principal  two  places  to  the  left, 

II.  To  find  the  interest  for  1  mo.  at  1  %  : — Divide  the  interest 
for  1  year  by  12. 

Ill  To  find  the  interest  for  any  time  at  1  %  :  —  Multiply  the 
interest  for  1  month  by  the  given  time  expressed  in  months  and 
tenths  of  a  month. 

lY.  To  find  the  interest  at  any  rate  %  :  —  Multiply  the  interest 
at  \  ^0  for  the  given  time  by  the  given  rate. 

Contractions.  After  removing  the  separatrix  in  the  principal  two 
places  to  the  left,  the  result  may  be  regarded  either  as  the  interest 
on  the  given  principal  for  12  months  at  1  per  cent.,  or  for  1  month  at 
12  per  cent.  If  we  regard  it  as  for  1  month  at  12  per  cent.,  and  if 
the  given  rate  be  an  aliquot  part  of  12  per  cent.,  the  interest  on  the 


310  PERCENTAGE. 

given  principal  for  1  month  may  readily  be  found,  by  taking  such  an 
aliquot  part  of  the  interest  for  1  month  as  the  given  rate  is  part  of  12 
per  cent.     Thus, 

To  find  the  interest  for  1  month  at  6  per  cent.,  remove  the  separa- 
trix  two  places  to  the  left,  and  divide  \)y  2. 

To  find  it  at  3  per  cent.,  proceed  as  before,  and  divide  by  4 ;  at  4 
per  cent.,  divide  by  3 ;  at  2  per  cent.,  divide  by  6,  etc. 

SIX   PER    CENT,    METHOD.* 

5410.  By  referring  to  «S3t>  it  will  be  seen  that  the  legal  rate 
of  interest  in  -^2  States  is  6  per  cent.  This  is  a  sufficient  reason 
for  introducing  the  following  brief  method  into  this  work : 

Analysis.     At  6  J/^  per  annum  the  interest  on  $1 

For  12  months is  $.06. 

*'      2monihs(i%  =  Jofl2mo.) "    .01. 

'*      1  month,  or  30  days  (y^  of  12  mo.)  "     .0OJ  =  $.005  (^j  of  $.06). 

"      6  days  (I  of  30  da.) *'     .001. 

**      1     "     (i  of  6  da.  =  jV  of  30  da.)  "     .OOOJ. 
Hence  we  conclude  that, 

1st.  The  interest  on  $1  is  $.005  per  month,  or  §.01  for  every 
2  months; 

2d.  The  interest  on  §1  is  g.OOOi  per  day,  or  ?.001  for  every  6 
days. 

From  these  principles  we  deduce  the 

lluLE.  I.  To  find  the  rate: —  Call  ever?/  year  §.06,  every  2 
montlu  $.01,  every  6  days  $.001,  and  any  less  number  of  days 
sixths  of  1  niill. 

II.  To  find  the  interest :  —  Multiply  the  jprincijyal  by  the  rate. 

Notes. — 1.  To  find  the  interest  at  any  other  rate  ^  by  this  method,  first  find 
it  at  6  ^ot  and  then  increase  or  c'lmin'sh  the  result  by  as  many  sixths  of  itself  as 
the  given  rate  is  units  greater  or  lc*i«s  than  G  ^.  Thus,  for  7  J^  add  ^,  fur  4  ^ 
subtract  J,  etc. 

2.  The  interest  of  $10  for  6  days,  or  of  $1  for  60  days,  is  $.01.  Therefore,  if  tho 
principal  be  less  than^  $10  and  the  time  less  than  6  days,  or  the  principal  less 
than  $1  and  the  time  less  than  60  days,  the  interest  will  be  less  than  $.01,  and 
may  be  disregarded. 

3.  Since  the  interest  of  $1  for  60  days  is  $.01,  the  interest  of  $1  for  any  num- 

*  This  method  of  finding  the  interest  on  $1  by  inspection  was  first  published 
in  The  Scholar's  Arithmetic,  by  Daniel  Adams,  M.  D.,  in  ISOl,  and  from  its 
simplicity  it  has  come  into  very  general  use. 


SIMPLE  INTEREST.  311 

ber  of  days  is  as  many  cents  as  60  is  contained  times  in  the  number  of  days. 
Therefore,  if  any  principal  be  multiplied  by  the  number  of  days  in  any  given 
number  of  months  and  dnys,  and  the  product  divided  by  60,  the  result  will  be 
the  interest  in  cents.  That  is,  Mnltiplji  the  pritio'jxrl  bij  the  uiinihvr  of  dnj/H, 
divide  the  pntdiict  by  60,  aiid  point  off  two  decimal.  pfficen  in  the  quotient.  The 
result  will  be  the  interest  iu  the  name  denominulion  as  the  principal, 

EXAMPLES    FOR   PRACTICE. 

What  is  the  interest  on  the  following  sums  for  the  times  given, 
at  6  per  cent.  ? 

1.  $325  ror  3  years.  Ans.  $58.50, 

2.  $1600  tor  1  yr.  3  mo.  Ans.   $120. 

3.  $36.84  for  5  mo. 

4.  $35.14  for  2  yr.  9  mo.  15  da. 

5.  $217.15  for  3  yr.  10  mo.  1  da.  Ans,  $49.98  +  . 

6.  $721.53  for  4  yr.  1  mo.  18  da. 

7.  $15,125  for  15  mo,  17  da.  Ans,  $1.17+. 
On  the  following  at  7  per  cent.  ? 

8.  ^2000  for  5  yr.  6  mo. 

9.  $1436.59  for  2  yr.  5  mo.  18  da.  Ans.  $248,051  +  . 

10.  $224.14  for  8  mo.  13  da.  Ans.  $11,026. 

11.  $100.25  for  63  da.  Ans.  S1.228+. 

12.  $600  for  24. da. 

13.  $520  for  5  yr.  11  mo.  29  da.  Ans.  $218,298. 

14.  $710.01  for  3  yr.  11  mo.  8  da. 
On  the  following  at  5  per  cent.  ? 

15.  $48,255  for  5  yr. 

16.  $750  for  l*yr.  3  mo. 

17.  $647,654  for  4  yr.  10  mo.  20  da.        Ans,  $158,315  +  . 

18.  $12850  for  90  da. 

19.  $2500  for  ^  mo.  20  da.  Ans.  $79.86. 

20.  $850.25  for  8  mo. 

I     21.  $48.25  for  1  yr.  2  mo.  17  da.  Ans.  $2,928  +  . 
On  the  following  at  8  per  cent.  ? 

22.  $2964.12  for  11  mo.  Ans.  $217,368  + 

23.  $725.50  for  150  da. 

24.  $360  for  2  yr  6  mo.  12  da. 

25.  $j600  for  3  yr.  2  mo.  17  da.  Ans,  $154.266f. 


312  PERCENTAGE. 

26.  $1700  for  28  da.  Ans.  $10-58-. 
On  the  following  at  10  per  cent.? 

27.  §3045.20  for  7  mo.  15  da.  '  '    Ans,  §190.32-f . 

28.  $1247.375  for  2  yr.  26  da.  Ans.  $258.48+. 

29.  $2450  for  60  da. 

30.  $375,875  for  3  mo.  22  da. 

31.  $5000  for  10  da. 

32.  $127.65  for  1  yr.  11  mo.  3  da.  Ans.  $24,572. 

33.  What  is  the  interest  of  $155.49  for  3  mo.,  at  6}  per  cent.  ? 

34.  What  is  the  interest  of  $970.99  for  6  mo.,  at  5i  per  cent.  ? 

35.  What  is  the  amount  of  $350.50  for  2  yr.  10  mo.,  at  7  per 
cent.?  Ans.  $120.01+. 

36.  What  is  the  interest  of  $95,008  for  3  mo.  24  da.,  at  4i  per 
cent.?  Ans.  $1,353+. 

37.  What  is  the  amount  of  $145.20  for  1  yr.  9  mo.  27  da.,  at 
12i  per  cent.?  ^«s.  $178.32375. 

38.  What  is  the  amount  of  $215.34  for  4  yr.  6  mo.,  at  3  J  per 
cent.?  Ans.  $U9.256+. 

39.  What  is  the  amount  of  $5000  for  20  da.,  at  7  per  cent.  ? 

40.  What  is  the  amount  of  $16941.20  for  1  yr.  7  mo.  28  da., 
at  4i  per  cent.  ?  Ans.  $18277.91—. 

41  If  $1756.75  be  placed  at  interest  June  29,  1860,  what 
amount  will  be  due  Feb.  12,  1863,  at  7  %  ? 

42.  If  a  loan  of  $3155.49  be  made  Aug.  15,  18§8,  at  6  per 
cent.,  what  amount  will  be  due  May  1,  1866,  no  interest  having 
been  paid? 

43.  How  mucb  is  the  interest  on  a  note  for  $257.81,  dated 
March  1,  1859,  and  payable  July  16,  1861,  at  7  %  ? 

44.  A  person  borrows  $3754.45,  being  the  property  of  a  minor 
who  is  15  yr.  3  mo.  20  da.  old.  He  retains  it  until  the  owner  is 
21  years  old.  How  much  money  will  then  be  due  at  6  %  simple 
interest?  Ans.  $5037.22+. 

45.  If  a  person  borrow  $7500  in  Boston  and  lend  it  in  Wis- 
consin, how  much  does  he  gain  in  a  year  ? 

46.  A  man  sold  a  piece  of  property  for  $11320;  the  terms  were 
$3200  in  cash  on  delivery,  $3500  in  6  mo.,  $2500  in  10  mo.,  and 


SIMPLE  INTEREST.  313 

the  remainder  in  1  yr.  3  mo.,  with  7  %  interest;  what  was  the 
whole  amount  paid  ?  Ans.  $11773.83  J. 

47.  May  10,  1859,  I  borrowed  $3840,  with  which  I  purchased 
Sour  at  $5.70  a  barrel.  June  21, 1860, 1  sold  the  flour  for  $6.62} 
a  barrel,  cash.  How  much  did  I  gain  by  the  transaction,  interest 
being  reckoned  at  6  %  ? 

48.  If  a  man  borrow  $15000  in  New  York,  and  lend  it  in 
Ohio,  how  much  will  he  lose  in  146  days,  reckoning  360  days  to 
the  year  in  the  former  transaction,  and  365  days  in  the  latter  ? 

49.  Hubbard  &  Northrop  bought  bills  of  dry  goods  of  Bowen, 
McNamee  &  Co.,  New  York,  as  follows,  viz. :  July  15,  1860, 
81250;  Oct.  4,  1860,  $3540.84;  Dec.  1,  1860,  $575;  and  Jan. 
24,  1861,  $816.90.  They  bought  on  time,  paying  legal  interest; 
how  much  was  the  whole  amount  of  their  indebtedness,  March  1, 
1861? 

50.  A  broker  allows  6  per  cent,  per  annum  on  all  moneys  de- 
posited with  him.  If  on  an  average  he  lend  out  every  $100  re- 
ceived on  deposit  11  times  during  the  year,  for  33  days  each 
time  at  2  %  a  month,  how  much  does  he  gain  by  interest  on 
$1000?  A71S.  $182. 

51.  A  man,  engaged  in  business  with  a  capital  of  $21840,  is 
making  12}  per  cent,  per  annum  on  his  capital;  but  on  account  of 
ill  health  he  quits  hi«  business,  and  loans  his  money  at  7i  %. 
How  much  does  he  lose  in  2  yr.  5  mo.  10  da.  by  the  change  ? 

Ans.  S2535.86f. 

52.  A  speculator  wishing  to  purchase  a  tract  of  land  containing 
450  acres  at  $27.50  an  acre,  borrows  th3  money  at  5}  per  cent. 
At  the  end  of  4  yr.  il  mo.  20  da.  he  sells  |  of  the  land  at  $34 
an  acre,  and  the  remainder  at  $32.55  an  acre.  How  much  does 
he  lose  by  the  transaction  ? 

53.  Bought  4500  bushels  of  wheat  at  $1.12}  a  bushel,  payable 
in  6  months;  I  immediately  realized  for  it  $1.06  a  bushel,  cash, 
and  put  the  money  at  interest  at  10  per  cent.  At  the  end  of  the 
6  months  I  paid  for  the  wheat ;  did  I  gain  or  lose  by  the  transac- 
tion, And  how  much  ? 

1'         27 


814  PERCENTAGE. 

TARTIAL  PAYMENTS  OR  INDORSEMENTS. 

«541:9.  A  Partial  Payment  is  payment  in  part  of  a  note,  bond, 
or  oilier  obligation. 

41 4S.  An  Indorsement  is  an  acknowledgment  written  on  the 
back  of  an  obligation,  stating  the  time  and  amount  of  a  partial 
payment  made  on  the  obligation. 

^•J.3.  To  secure  uniformity  in  the  method  of  computing  in- 
terest where  partial  payments  have  been  made,  the  Supreme  Court 
of  the  United  States  has  decided  that, 

I.  ^'  The  rule  for  casting  interest  when  partial  payments  have 
been  made,  is  to  apply  the  payment,  in  the  first  place,  to  the  dis- 
charge ot  the  interest  then  due. 

II.  "  If  the  payment  exceeds  the  interest  the  surplus  goes  to- 
wards discharging  the  principal,  and  the  subsequent  interest  is  to 
be  computed  on  the  balance  of  the  principal  remaining  due. 

III.  '^  If  the  payment  be  less  than  the  interest  the  surplus  of 
interest  miist  not  be  taken  to  augment  the  principal,  but  the  inte- 
rest continues  on  the  former  principal  until  the  period  when  the 
payments,  taken  together,  exceed  the  interest  due,  and  then  the 
surplus  is  to  be  applied  towards  discharging  the  principal,  and  the 
interest  is  to  be  computed  on  the  balance  as  aforesaid.'^ — Decision 
of  Chancellor  Kent. 

This  decision  has  been  adopted  by  nearly  all  the  States  of  the 
Union,  the  only  prominent  exceptions  being  Connecticut,  Ver- 
mont* and  New  Hampshire.  We  therefore  present  the  method 
prescribed  by  this  decision  as  the 

United  States  Kule. 

I.  Find  the  amount  of  the  given  principal  to  tlie  tirm  of  Hie 
first  iiaymentj  and  if  this  payment  exceed  the  interest  then  diie^ 
TMhiract  it  from  the  amount  obtained,  and  treat  the  remainder  as  a 
new  j)ri'nrj'pal. 

II.  But  if  the  interest  he  greater  than  any  payment ^  compute  the 
interest  on  the  same  principal  to  a  time  when  the  sum  of  the  jjcy- 
ments  shall  equal  or  exceed  the  interest  due,  and  subtract  the  sum 


PARTIAL  PAYMENTS.  315 

oj  the  'payments  from  the  amount  of  the  principal ;  the  remainder 
will  form  a  new  principal ^  with  which  proceed  as  he/ore, 

EXAMPLES   FOR   PRACTICE. 

flOOO.  Buffalo,  N   Y.,  May  15  1856. 

1.  Two  years  after  date  I  promise  to  pay  to  David  Hudson,  or 
order,  one  thousand  dollars,  with  interest,  for  value  received. 

Henry  Burr. 
On  this  note  were  indorsed  the  following  payments : 

Sept.  20,  1857,  received, $150.60 

Oct.    25,1859,      ''        200.90 

July    11,  1861,       "        75.20 

Sept.  20,  1862,      " 112.10 

Dec.      5,  1863,       "        105. 

What  remained  due  May  20,  1864  ? 

OPERATION. 

Principal  on  interest  from  May  15,  1856, $1000 

Interest  to  Sept.  20,  1857,  1  yr.  4  mo  5  da., 94.31 

Amount, $1094.31 

1st  Payment,  Sept  20,  1857, 150.60 

Remainder  for  a  new  principal, $943.71 

Interest  from  1st  paym't  to  Oct.  25, 1859,  2  yr.  1  mo.  5  da.,  138.54 

Amount, $1082.25 

2d  Payment,  Oct.  25,  1859, 200.90 

Remainder  for  a  new  principal, $881.35 

Interest  from  2d  paym't  to  Dec.  5, 1863,  4  yr.  1  mo.  10  da.,         253.63 

Amount, $1134.98 

3d  Payment,  less  than  interest  due, $75.20 

4th         "  112.11 

Sura  of  3d  and  4th  payments,  less  than  interest  due,  $187.31 
5th  payment, 105.00 

Sums  of  3d,  4th,  and  5th  payments, 292.31 

Remainder  for  new  principal,. $842.67 

Interest  to  May  20,  1864,  5  mo.  15  da., 27.04 

Balance  due  May  20,  1864, $869.71 


316  PERCENTAGE. 


$^^0^-  Richmond,  Va.,  Oct.  15, 1859. 

2.  One  year  after  date  we  promise  to  pay  Jatnes  Peterson,  or 
order,  twelve  hundred  dollars,  for  value  received,  with  interest. 

Wilder  k  Son. 

Indorsed  as  follows:    Oct.  15,  1860,  $1000;  April  15,  1861, 
§200.    How  much  remained  due  Oct.  15, 1861  ?     Ans,  ?82.56. 


^^50//^.  Boston,  June  10,  1855. 

3.  Eighteen  months  after  date  I  promise  to  pay  Crosby,  Nich- 
ols &  Co.,  or  order,  eight  hundred  fifty  and  j'y^  dollars,  with 
interest,  for  value  received.  0.  L.  Sanborn. 

Indorsed  as  follows:  March  4,  1856,  $210.93;  July  9,  1857, 
$140;  Feb.  20,  1858,  $178;  May  5,  1859,  $154.30;  Jan.  17, 
1860,  $259,45.     How  much  was  due  Oct.  24,  1861  ? 


$2^j¥Tr  Savannah,  Ga.,  Sept.  4, 1860. 

4.  Six  months  after  date  I  promise  to  pay  John  Rogers,  or 
order,  three  hundred  eighty-four  and  -j^^^^  dollars,  for  value  re- 
ceived, with  interest.  \Vm.  Jenkins. 

This  note  was  settled  Jan.  1,  1862,  one  payment  of  $126.50 
having  been  made  Oct.  20,  1861 ;  Ibow  much  was  due  at  the  time 
of  Settlement  ? 


^^^"'^-  New  Orleans,  March  6,  1857. 

5  On  demand  we  promise  to  pay  Evans  &  Hart,  or  order,  three 
thousand  four  hundred  seventy-five  dollafrs,  for  value  received,  with 
interest.  Davis  &  Brother. 

Indorsed  as  follows-  June  1,  1857,  $1247.60;  Sept  10,  1857, 
$1400.     How  much  was  due  Jan.  31,  1858  ? 

6.  A  gentleman  gave  a  mortgage  on  his  estate  for  $9750,  dated 
April  1,  1860,  to  be  paid  in  5  years,  with  annual  interest  after  9 
months  on  all  unpaid  balances,  at  10  per  cent.  Six  months  from 
date  he  paid  $846.50;  Oct.  20, 1862,  $2500;  July  3, 1863,  $1500; 
Jan.  1,  1864,  $500;  how  much  was  due  at  the  expiration  of  the 
given  time? 


PARTIAL  PAYMENTS.  317 


g5^Q»  Philadelphia,  Feb.  1,  1861. 

7.  For  value  received,  I  promise  to  pay  J.  B.  Lippincott  &  Co., 
or  order,  five  hundred  dollars  three  months  after  date,  with  interest. 

James  Monroe. 

Indorsed  as  follows:  May  1,  1861,  $10;  Nov.  14,  1861,  $8; 
April  1,  1862,  $12;  May  1,  1862,  $30.  How  much  was  due 
Sept.  16,  1862?  Ans.  $455.57+. 

544.    Connecticut  Rule. 

I.  Payments  made  one  year  or  more  from  the  time  the  interest  ■ 
commenced  J  or  from  another  payment  ^  and  'payments,  less  lluin  the 
interest  due,  are  treated  according  to  the  United  Sfates  rule. 

II.  Payments  exceeding  the  interest  due  and  made  within  one 
year  from  the  time  interest  commenced ,  or  from  a  former  paymc7it, 
shall  draw  interest  for  the  balance  of  the  year^  p)rovidf'd  the  interval 
does  not  extend  hey  and  the  settlment^  and  the  amount  mwt  be  sub- 
tracted from  the  amount  of  the  principal  for  otie  year;  the  re- 
maijider  will  be  the  new  principal. 

III.  If  the  year  extend  beyond  the  settlemrnt^  then  find  the 
amount  of  the  payment  to  the  day  of  settlement,  and  subtract  it 
from  the  amount  of  the  principal  to  that  day  ;  the  remainder  will 
be  the  sum  due. 

•145.  A  note  containing  «-  promise  to  pay  interest  annually 
is  not  considered  in  law  a  contract  for  any  thing  more  than  simple 
interest  on  the  principal.  Fir  partial  payments  on  such  notes 
the  following  is  the 

Vermont  Rule. 

I.  Find  the  amount  of  the  principal  from  the  time  interest  com- 
menced to  the  time  of  settlement. 

II.  Fiiid  the  amount  of  each  payment  from  the  time  it  icas  made 
to  the  time  of  settlement. 

III.  Subtract  the  sum  of  the  amounts  of  the  payments  from  the 
amount  of  the  pi'incipal ;    the  rc-niainder  will  be  the  sum  due. 
_NoTE. — This  rule  is  in  quite  extensive  use  among  merchants  and  others. 

27* 


318  PERCENTAGE. 

S4rG«  In  New  Hampshire  interest  is  allowed  on  the  annual 
interest  if  not  paid  when  due,  in  the  nature  of  damages  for  its 
detention;  and  if  payments  are  made  he/ore  one  year's  interest 
has  accrued,  interest  must  be  allowed  on  such  payments  for  the 
balance  of  the  year.     Hence  the  following 

New  Hampshire  Rule. 

I.  Find  the  amount  of  the  principal  for  one  year^  and  deduct 
from  it  the  amount  of  each  payment  of  that  year,  from  the  time 
it  was  made  up  to  the  end  of  the  year ;  the  remainder  will  he  a 
new  principal,  with  which  proceed  as  before. 

II.  If  the  settlement  occur  less  than  a  year  from  the  last  annual 
term  of  interest,  make  the  last  term^f  interest  a  part  of  a  year, 
accordingly, 

EXAMPLES   FOR   PRACTICE. 


^IQQQ-  New  Haven,  Conn,  Feb.  1,  185G. 

1.  Two  years  after  date,  for  value  received,  I  promise  to  pay  to 
Peck  &  Bliss,  or  order,  one  thousand  dollars  with  interest. 

John  Cornwall. 

Indorsed  as  follows:  April  1,  1857,  ^80;  Aug.  1,  1857,  $30; 
Oct.  1, 1858,  $10 ;  Dec.  1, 1858,  $600 ;  May  1, 1859,  $200.  How 
much  was  due  Oct.  1,  1859  ?  Ans,  $266.38. 


$2000. 


Burlington,  Yt.,  May  10,  1858. 


2.  For  value  received,  I  promise  to  pay  David  Camp,  or  order, 
two  thousand  dollars,  on  demand,  with  interest  annually. 

Richard  Thomas. 
On  this  note  were  indorsed  the  following  payments :  March  10, 
1859,  $800;  May  10,  1860,  $400;  Sept.  10,  1861,  $300.     How 
much  was  due  Jan.  10,  1863  ? 

3.  How  much  would  be  due  on  the  above  note,  computing  by 
the  Connecticut  rule  ?  Ans.  $831.58. 

4.  How  much,  computing  by  the  New  Hampshire  rule?     By 
the  United  States  rule  ? 


.        f  N.  H.  rule,  $833.21; 
""''  1 U.  S.     "      $831.90. 


SAVINGS    BANK    ACCOUNTS.  319 

SAVINGS   BANK    ACCOUNTS. 

547.  Saving's  Banks  are  institutions  intended  to  receive  in 
trust  or  on  deposit,  small  sums  of  money,  generally  the  surplus 
earnings  of  laborers,  and  to  return  the  same  with  a  moderate  interest 
at  a  future  time. 

548.  It  is  the  custom  of  all  savings  hanks  to  add  to  each 
depositor's  account,  at  the  end  of  a  certain  fixed  term,  the  interest 
due  on  his  deposits  according  to  some  general  regulation  for  allow- 
ing interest.  The  interest  term  with  some  savings  banks  is  6 
months,  with  some  3  months,  and  with  some  1  month. 

540.  A  savings  bank  furnishes  each  depositor  with  r.  book, 
in  which  is  recorded  from  time  to  time  the  sums  deposited  and 
the  sums  drawn  out.  The  Dr.  side  of  such  an  account  shews  the 
deposits,  and  the  Cr.  side  the  depositor's  checks  or  drafts.  In  the 
settlement,  interest  is  never  allowed  on  any  sum,  which  has  not 
been  on  deposit  for  a  full  interest  term.  Hence,  to  find  the 
amount  due  on  any  depositor's  account,  we  have  the  following 

Rule.  At  the  end  of  each  term^  add  to  the  hcdance  of  the 
account  one  term^s  interest  on  the  smallest  balance  on  deposit  at  any 
one  time  durincj  that  term  j  the  final  balance  thus  obtained  loill  be 
the  sum  due. 

Notes.  —  1.  It  will  be  seen  that  by  this  rule  no  interest  is  allowod  for 
money  on  deposit  during  a  partial  term,  whether  the  period  be  the  first  or  tlio 
last  part  of  the  term. 

2.  An  exception  to  this  general  rule  occurs  in  the  practice  of  some  of  tlio 
savings  banks  of  New  York  city.  In  these,  the  interest  term  is  G  moiitlis,  and 
the  depositor  is  allowed  not  only  the  full  term's  interest  on  the  smallest.  Inlanco, 
but  a  half  term's  interest  on  any  deposit,  or  portion  of  a  deposit  made  during 
the  first  3  months  of  the  term,  and  not  dratcn  out  duriuij  ant/  subsequent  2)art  of 
the  term. 

EXAMPLES    FOR   PRACTICE. 

1.  What  will  be  due  April  20,  1860,  on  the"  following  account, 
interest  being  allowed  quarterly  at  6  per  cent,  per  annum,  tho 
terms  commencing  Jan.  1,  April  1,  July  1,  and  Oct.  1  ? 

Dr.       Savings  Bank  in  account  with  James  Taylor.        Cr. 

1858,  Jan.    12,... S75         1858,  March  5, ^-30 

"       May  10, 150  "       Aug.  16, 50 

.  '^       Sept     1, 20  "       Dec.     1, 48 

1859,  Feb.  16, 130 


320 


PERCENTAGE. 


OPERATION. 


Deposit,  Jan.  12,  1858, $75 

Draft,  March  5,       "     _30 

Balance,  Apr.  1,  1860, $45 

Deposit,  May  10,  1858, 150 

Int.  on  $45,  for  3  mo 68 

Balance,  July  1,  1860, $1-95.68 

Draft,  Aug.  16,  1858, 50 

Least  balance  during  the  current  term, $145.68 

Deposit,  Sept.  1,  1858, 20.00 

Int.  on  $145.68,  for  3  mo 2.19 

Balance,  Oct.  1,  1858, $167.87 

Draft,  Dec.  1,  1858, 48 

Least  balance  during  the  current  term, 119  87 

Int.  on  $119.87,  for  3  mo 1.80 

Balance,  Jan.  1,  1860, $121.67 

Deposit,  Feb.  16,1860, ^ 130.00 

Int.  on  $121.67,  for  3  mo 1.83 


Note.- 
Jan.  1. 


Bal.  due  after  Apr.  1,  1860, $253.50  ^n^. 

-In  the  following  examples  the  terms  commence  with  the  year,  or  on 


2.  Allowing  interest  monthly  at  6  %  per  annum,  what  sum 
will  be  due  Sept.  1,  1860,  on  the  book  of  a  savings  bank  having 
the  following  entries  ? 


Dr. 


Bay  State  Savings  Institution,  in  account  with  Jane  Ladd. 


Cr. 


1860. 

Jan. 

3 

«( 

8 

i< 

20 

Feb. 

20 

27 

March 

6 

29 

April 
May 

25 

7 

30 

July 
Aug. 

28 
3 

u 

26 

To  cash, 


"  check, 

"  cash, 

"  check, 

"  cash, 

"  draft, 

H  ii 

"  cash, 

"  check, 

*'  cash, 


5 

75 

13 

45 

7 

60 

16 

45 

8 

40 

14 

65 

7 

98 

3 

49 

26 

60 

45 

79 

15 

68 

18 

45 

4 

60 

1860. 
Jan. 
Feb. 
March 
April 
June 


Aug. 


28 
7 
20 
11 
3 
12 
20 
17 


By  check, 


"   draft, 
"   check, 


k, 

5 

00 

8 

48 

10 

00 

12 

76 

3 

96 

10 

48 

, 

17 

48 

^, 

5 

64 

Ans.  $116.87. 


3.  Interest  at  7  %,  allowed  quarterly,  how  much  was  due  April 
4,  1860,  on  the  following  savings  bank  account  ? 


COMPOUND  INTEREST. 


321 


Detroit  Savings  Institution,  in  account  tvith  E.  L.  SeJclen, 


Cr 


1S<,9. 

J  n. 

1 

M.-.rch 

12 

June 

20 

Aujr. 

3 

isco. 

Jan. 

25 

To  cash, 


}^.9. 

47 

ro 

May 

1-24 

30 

Oct. 

13) 

6(5 

Nov. 

08 

75 

1     l»ec. 

ICO 

80 

1 

12     J     By  check, 


If) 

28 


50  !  36 
25  i  78 
36  j  48 
12      50 


Ans.  S423.22. 
4.  How  mnch  was  due  Jan.  1,  1860,  on  the  following  account, 
allowing  interest  semi-annually,  at  6  %  per  annum  ? 

Irvings  Savings  Institution,  in  account  with  James  Taylor, 


1858. 

1858. 

Junu 

4 

To  cash, 

175 

S.'pt. 

14 

Nov. 

1 

u        a 

150 

1859. 

185S). 

July 

25 

Feb. 

24 

''    draft. 

200 

Dec. 

3 

Sept. 

10 

«    check, 

66 

By  check, 


120 
80 


'   Ans.  $337.02. 
5.  Interest  at  5  %,  allowed  according  to  Note  2,  how  much  was 
due,  Jan.  1,  1860,  on  the  book  of  a  savings  bank  in  the  city  of 
New  York,  having  the  following  entries  ? 


Dr. 


Sixpenny  Savings  Bank,  in  account  with  William  Gallup, 


1858. 

1     1858. 

Jau. 

1 

To  check, 

30 

60 

Sept. 

16 

By  check. 

March 

17 

a         .i 

26 

38 

1   1859. 

Aujr. 

1 

"    cash, 

84 

72 

1   Jan. 

27 

«       (t 

Ib59. 

March 

1 

«       (( 

June 

11 

"   draft, 

60 

00 

Nov. 

1(5 

"   cubh, 

40 

78 

Ans,  §179.10. 


COMPOUND  INTEREST. 

•>«^0«  Compound  Interest  is  interest  on  both  principal  and 
interest,  when  the  interest  is  not  paid  when  due. 

NoTK. —  The  ."imple  interest  m  ly  be  addled  to  the  principal  nnnnally,  Rp»ni- 
jniiiually,  or  quarterly,  as  the  partiea  may  agree  -,  but  the  taking  of  compui  nd 
interest  in  not  legal. 

1  What  is  the  compound  interest  of  $640  for  4  years,  ^i  5 
per  cent.  ? 

V 


822  PERCENTAGE. 

OPERATION. 

$610  Principal  for  1st    year, 

$340  X  1.05  =  $372  ''         "   2d       '' 

$872  X  1.05  =  $705.60     "    "   3d   " 

$705.60  X  1.05  =  $740.88__    "    "  4th  '' 

§740.88  X  1.05  ==:  S777.924  Amount     «   4  years, 
640.  Given  principal, 

$137,924    Compound  interest. 
This  illustration  is  sufficient  to  establish  the  folloAving 
KuLE.     I.   Find  the  amount  of  the  given  'principal  at  the  given 
rate  for  one  year^  and  make  it  the  principal  for  the  second  year, 

II.  Find  the  amount  of  this  new  principal,  and  mahe  it  the 
principal  for  the  third  year,  and  so  continue  to  do  for  the  given 
number  of  yrars, 

III.  Subtract  the  given  principal  from  the  last  amount ;  the  re- 
mainder will  be  the  compound  interest. 

Notes. — 1.  When  the  interest  is  payable  femi-annually  or  quarterly,  find  the 
amount  of  the  given  principal  for  the  first  interval,  and  make  it  the  principal 
for  the  second  interval,  proceeding  in  all  respects  as  when  the  interest  is  payable 
yearly. 

2.  When  the  time  contains  years,  months,  and  days,  find  the  amount  for  the 
years,  upon  which  compute  the  interest  for  the  months  and  days,  and  add  it  to 
the  last  amount,  before  subtracting. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  compound  interest  of  $750  for  4  years  at  6  per 
cent.?  Ans.  $196.86- 

2.  What  will  $250  amount  to  in  3  years  at  7  per  cent,  compound 
interest?  Ans.  $306.26. 

3.  At  7  per  cent,  interest,  compounded  semi-annually,  what 
debt  will  $1475.50  discharge  in  2i  years?  Ans.  $1752.43. 

4.  Find  the  compound  interest  of  $376  for  3  yr.  8  mo.  15  da.;, 
at  6  per  cent,  per  annum.  J^ns.  $90.84. 

^«il«  A  more  expeditious  method  of  computing  compound 
interest  than  the  preceding  is  by  the  use  of  the  compound  interest 
table  on  the  following  page. 


COMPOUND  INTEREST. 


823 


TABLE, 

Showing  the  amount  of  $1,  or  <£!,  at  2J,  3,  3J,  4,  5,  G,  7,  and  8  '^.er 

cent.,  compound  interest,  for  any  number  of  years  from  1  to  40. 


Years 
1 

2Kpe'-ct. 

3  percent. 

sT^perct. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

1.025000 

1.030000 

1.035000 

1.040000 

1.050000 

l.OGOOOO 

1.070000 

l.OSOOOO 

2 

1.050625 

1.060900 

1  071225 

1.081600 

1.102500 

1.123600 

1.1449U0 

1.166-100 

3 

1.076891 

1.092727 

1.108718 

1.124864 

1.157625 

1.191016 

1.225043 

1.259712 

4 

1.103813 

1.125509 

1.147523 

1.169859 

1.215506 

1.262477 

1.310796 

1.3G0489 

5 

1.131408 

1.159274 

1.187686 

1.216653 

1.276282 

1.338226 

1.402552 

1.469328 

6 

1.159693 

1.194052 

1.229255 

1.265319 

1.340096 

1.418519 

1.500730 

1.586874 

7 

1.188686 

1.229874 

1.272279 

1.315932 

1.407100 

1.503630 

1.605782 

1.713824 

8 

1.218403 

1.266770 

1.316809 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9 

1.248863 

1.304773 

1.362897 

1.423312 

1.551328 

1.689479 

1.8G84.>9 

1.999005 

TkJ 

1.280085 

1.343916 

1.410599 

1.480244 

1.628885 

1.790848 

1.967151 

2.158925 

11 

1.312087 

1.384234 

1.459970 

1.539454 

1.710339 

1.898299 

2.104852 

2.331639 

12 

1.344889 

1.425761 

1.511069 

1.601032 

1.795856 

2.012197 

2.252192 

2.518170 

13 

1.378511 

1.468534 

1.563956 

1.665074 

1.885649 

2.132928 

2.409845 

2.719624 

14 

1.412974 

1.512590 

1.618695 

1.731676 

1.979932 

2.260904 

2.578534 

2.937194 

15 

1.448298 

1.557967 

1.675349 

1.800944 

2.078928 

2.396558 

2.759032 

3.172169 

16 

1.484506 

1.604706 

1.733986 

1.872981 

2.182875 

2.540352 

2.952164 

3.425943 

17 

1.521618 

1.652848 

1.794676 

1.947901 

2.292018 

2.692773 

3.15S815 

3.700018 

18 

1.559659 

1.702433 

1.857489 

2.025817 

2.406619 

2.854339 

3.379932 

3.99C020 

19 

1.598650 

1.753506 

1.922501 

2.106849 

2.526950 

3.025600 

3.616528 

4.315701 

20 

1.63S616 

1.806111 

1.989789 

2.191123 

2.653298 

3.207136 

3.8G9CS5 

4.660957 

21 

1.679582 

1.860295 

2.059431 

2.278768 

2.785963 

3.399564 

4.140562 

5.033834 

22 

1.721571 

1.916103 

2.131512 

2.369919 

2.925261 

3.603537 

4.4-30402 

5.43C540 

23 

1.764611 

1.973587 

2.206114 

2.464716 

3.071524 

3.819750 

4.740530 

5.871464 

24 

1.808726 

2.032794 

2.283328 

2.563304 

3.225100 

4.048935 

5.072.307 

6.341181 

25 

1.853944 

2.093778 

2.363245 

2.665836 

3.386355 

4.291871 

5.427433 

6.848475 

26 

1.900293 

2.156591 

2.445959 

2.772470 

3.555673 

4.549383 

5.807353 

7.396353 

27 

1.947800 

2.221289 

2.531567 

2.883369 

3.733456 

4.822346 

6.21C868 

7.98s^062 

28 

1.996495 

2.287928 

2.620172 

2.99S703 

3.920129 

5.111687 

6.648838 

8.627100 

29 

2.046407 

2.356566 

2.711878 

3.118651 

4.116136 

5.418388 

7.114257 

9.317275 

30 

2.097568 

2.427262 

2.806794 

3.243398 

4.321942 

5.743491 

7.612255 

10.062657 

31 

2.150007 

2.500080 

2.905031 

3.373133 

4.538040 

6.088101 

8.145113 

10.867C69 

32 

2.203757 

2..575083 

3.006708 

3.508059 

4.764942 

6.453387 

8.715271 

11.737083 

33 

2.258S51 

2.652335 

3.111942 

3.648381 

5.003189 

6,840590 

9.325340 

12.076050 

34 

2.315322 

2.731905 

3  220860 

3.794316 

5.253348 

7.251025 

9.97S114 

13.690134 

35 

2.373205 

2.813862 

3.333590 

3.946089 

5.516015 

7.686087 

10.676582 

14.785344 

36 

2.432535 

2.898278 

3.450266 

4.103933 

5.791816 

8.147252 

11.423942 

15.968172 

37 

2.493349 

2.985227 

3.571025 

4.2GS090 

6.081407 

8.6360S7 

12.223618 

17.245026 

38 

2.555682 

3.074783 

3.696011 

4.438S13 

6.-385477 

9.154252 

13.079271 

1 8.625276 

39 

2.619.=i74 

3.167027 

3.825372 

4.616366 

6704751 

9.703508 

13.994820 

20.115298 

40 

2  685064 

3.262038 

3.959260 

4.801021 

7.039989 

10.2'«5718 

14.974458 

21.724522 

324  PERCENTAGE. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  amount  of  $300  for  4  years  at  6  per  cent,  com- 
pound interest,payable  semi-annually  ? 

OPERATION.  Analysis.    The  amount  of  $1  at  6  per  cent., 

^1  26677  compound  interest  payable  semi-annually,  is 

300  the  same  as  the  amount  of  $1  at  3  per  cent., 

£^^0  OtToO  compound   interest    payable    annually.     We 

^      '  therefore  take,  from  the  table,  the  amount  of 

$1  for  8  years  at  3  per  cent.,  and  multiply  this  amount  by  the  given 

principal. 

2.  What  is  the  amount  of  $536.75  for  12  yr.  at  8  per  cent,  com- 
pound interest  ?  ^ws.  $1351.63. 

3.  What  sum  placed  at  simple  interest  for  2  yr.  9  mo.  12  da., 
at  7  per  cent.,  will  amount  to  the  same  as  $1275,  placed  at  com- 
pound interest  for  the  same  time  and  at  the  same  rate,  payable 
semi-annually?  Ans.  $1292.51 — . 

4.  At  8  per  cent,  interest  compounded  quarterly,  how  much 
will  $1840  amount  to  in  1  yr.  10  mo.  20  da.  ?     Ans.  $2137.06. 

5.  A  father  at  his  death  left  $15000  for  the  benefit  of  his  only 
son,  who  was  12  yr.  7  mo.  12  da.  old  when  the  money  was  de- 
posited; the  same  was  to  be  paid  to  him  when  he  should  be  21 
years  of  age,  together  with  7  per  cent,  interest  compou'uded  semi- 
annually.    How  much  was  the  amount  paid  him  ? 

6.  What  sum  of  money  will  amount  to  $2902.263  in  20  years, 
at  7  %  compound  interest  ?  Ans.  $750. 


PROBLEMS  IN  INTEREST. 
PROBLEM    I. 

SS3.   Given,  the  time,  rate  per  cent.,  and  interest,  to 
find  the  principal. 

1.  T7hat  sum  of  money  will  gainr  $87.42  in  4  years,  at  6  per 
cent.  ? 

OPERATION.  Analysis.     Since   $.24 

$.24,  interest  of  $1  for  4  years.  ^«  *^^  interest  of  $1  for  4 

$87.42  — -  .24  =  $364.25,  Ans,         years  at  6  per  cent.,  $87.42 

must  be  the  interest  of  as 


PROBLEMS  IN  INTEREST.  325 

many  dollars,  for  the  same  time  and  at  the  same  rate,  as  $.24  is  con-* 
tained  times  in  $87.42.  Dividing,  we  obtain  $364.25,  the  required 
principal.     Hence  the 

Rule.     Divide  the  given  interest  hy  the  interest  of  %\  for  the 
given  time  at  the  given  rate, 

EXAMPLES    FOR   PRACTICE. 

1.  What  sum  of  money,  invested  at  6 J  per  cent.,  will  produce 
$279,825  in  1  yr.  6  mo.?  Ans.  $2870. 

2.  What  sum  will  produce  $63.75  interest  in  6  mo.  24  da.  at 
7  J  per  cent.  ? 

3.  What  sum  will  produce  $12  J  interest  in  10  days  at  10  per 
cent.?  Alls.  $4500. 

4.  What  sum  must  be  invested  in  real  estate  paying  12  i  per 
cent,  profit  in  rents,  to  give  an  income  of  $3125  ? 

5.  What  is  the  value  of  a  house  and  lot  that  pays  a  profit  of  9i 
per  cent,  by  renting  it  at  $30  per  month  ? 

6.  What  sum  of  money,  put  at  interest  6  yr.  5  mo.  11  da.^  at 
7  per  cent.,  will  gain  $3159.14  ?  Ans.  $7000. 

7.  What  sum  of  money  will  produce  $69.67  in  2  yr.  9  mo.  at 
6  ^0  compound  interest  ?  Ans.  $400. 

8.  What   principal   at   6  %  compound   interest  will  produce 
$124.1624  in  1  yr.  6  mo.  15  da.  ?  Ans.  $1314.583. 

PROBLEM   II. 

^o3.   Given,  the  time,  rate  per  cent.,  and  amount,  to 

find  the  principaL 

1.   What  sum  of  money  in  2  years  6  months,  at  7  per  cent., 

will  araount.te  $136,535? 

OPERATION.  Analysis.    Since 

$1,175,  amount  of  $1  for  2  yr.  6  mo.  ^1-175  istheamount 

$136,535  -  1.175  =  $116.20,  Ans.  ^^  ^  V"'  ^  ^f'  ^ 

months,    at    7    per 

cent.,  $136,535  must  be  the  amount  of  as  many  dollars,  for  the  same 

time  and  at  the  same  rate,  as  $1,175  is  contained  times  in  $136,535. 

Dividing,  w©  obtain  $116.20,  the  required  principal.     Hence  the 

2^ 


326  PERCENTAGE. 

Rxjle.  Divide  flic  given  amoitnt  hy  the  amount  q/*  §1  for  the 
given  time  at  the  given  rate. 

EXAMPLES    FOR    TRACTICE. 

1.  \Yhat  principal  in  2  yr.  3  mo.  10  da.^  at  5  per  cent.,  will 
amount  to  $1893  61^  ?  Ans,  $1700. 

2.  A  note  which  had  run  3  yr.  5  mo.  12  da.  amounted  to 
$081,448,  at  6  per  cent. ;  how  much  was  the  face  of  the  note  ? 

3.  What  sum  put  at  interest  at  3  J  per  cent.,  for  10  yr.  2  mo., 
will  amount  to  $15660? 

4.  "What  is  the  interest  of  that  sum  for  2  yr.  8  mo.  29  da.,  at  7 
per  cent.,  which  at  the  same  time  and  rate,  will  amount  to 
$1568.97?  Ans.  $253,057+. 

5.  What  is  the  interest  of  that  sum  for  243  days  at  8  per  cent., 
which  at  the  same  time  and  rate,  will  amount  to  $11119.70  ? 

6.  What  principal  in  4  years  at  6  per  cent,  compound  interest, 
will  amount  to  $8644.62  ?  Ans.  $6847.34. 

7.  What  sum  put  at  compound  interest  will  amount  to  $26772.96, 
in  10  yr.  5  mo.,  at  6  per  cent.  ? 

Ans.  $14585.24. 

PROBLEM   III. 

554.  Given,  the  principal,  time,  and  interest,  to  find 
the  rate  per  cent. 

1.  I  received  $315  for  3  years*  interest  on  a  mortgage  of 
$1500;  what  was  the  rate  per  cent.  ? 

OPERATION.  Analysis.      Since 

^15^00  $^^  ^s  the  interest  on 

3  the    mortgage    for    3 

TTTT77    .        »     r.  ^    ^  years  at  1  per  cent., 

$45.00,  int.  for  3  yr.  at  1  %.  Ig^g  ^„^,  {^  ^^^^  .^_ 

$315  ~  $45  =  7  %,  A)is.  terest  on  the  mortgage 

for  the  same  time,  at 
as  many  times  1  per  cent,  as  $45  is  contained  times  in  $315.  Divid- 
ing, and  we  obtain  7,  the  required  rate  per  cent.     Hence  the 

Rule.  Divide  the  given  interest  hy  the  interest  on  the  principal 
for  the  given  time  at  1  per  cent. 


PROBLEMS  IN  INTEREST.  327 

EXAMPLES    FOR   PRACTICE. 

1.  If  I  loan  $750  at  simple  interest,  and  at  the  end  of  1  yr.  3 
mo.  receive  $796,874,  what  is  the  rate  per  cent.  ?  Ans.  5. 

2  If  I  pay  $10.58  for  the  use  of  $1700,  28  days,  what  is  the 
rate  of  interest?  Ans.  8-hper  cent. 

3.  Borrowed  $600,  and  at  the  end  of  9  yr.  6  mo.  returned 
$356.50 ;  what  was  the  rate  per  cent.  ? 

4.  A  man  invests  $7266.28,  which  gives  him  an  annual  income 
of  $744.7937;  what  rate  of  interest  does  he  receive  ? 

5.  If  C  buys  stock  at  30  per  cent,  discount,  and  every "^6  months 
receives  a  dividend  of  4  per  cent.,  what  annual  rate  of  interest 
does  he  receive  ?  Ans.  11|  per  cent. 

6.  At  what  rate  per  annum  of  simple  interest  will  any  sum  of 
money  double  itself  in  4,  6,  8,  and  10  years,  respectively  ? 

7.  At  what  rate  per  annum  of  simple  interest  will  any  sum 
triple  itself  in  2,  5,  7,  12,  and  20  years,  respectively  ? 

8.  A  house  that  rents  for  $760.50  per  annum,  cost  $7800 :  what 
%  does  it  pay  on  the  investment?  Ans.  9|  per  cent. 

9.  I  invest  $35680  in  a  business  that  pays  me  a  profit  of  $223  a 
month ;  what  annual  rate  of  interest  do  I  receive  ?     Ans.  7i  5^. 

PROBLEM   IV. 

535.  Given,  the  principal,  interest,  and  rate,  to  find 
the  time. 

1.  In  what  time  will  $924  gain  $151,536,  at  6  per  cent.? 

OPERATION.  ^    Analysis.      Since 

4924  $55.44  is  the  interest 

.06  of  $924  for  1  year  at 

$5144,  int.  of  $924  for  1  yr.  at  6  X.        ^  ^7  ^^'^^  ^^^^^^^^ 
'  J  /«>  jjmst  \yQ  the  interest 

$151,536  ^  $55.44  =  2.73  ^f  the  same  sum,  at 

2.73  yr.  =  2  yr.  8  mo.  24  da.,  Ans.         the    same    rate     per 

I  cent.,    for    as    many 

irs  as  $55.44  is  contained  times  in  $151,536,  which  is  2.73  times, 
ducing  the  mixed  decimal  to  its  equivalent  compound  number, 
have  2  years  8  months  24  days,  the  required  time.     Hence  the 


S28  PERCENTAGE. 

Rule.  Divide  the  given  interest  hi/  the  intereft  on  the  jirincijial 
for  1  year ;  the  quotient  will  he  the  required  time  in  years  and 
decimals, 

EXAMPLES    FOR   PRACTICE. 

1.  In  what  time  will  $273.51  amount  to  §312.864,  at  7  per 
cent.  ?  Ans,  2  yr.  20  da. 

2.  How  long  must  $650.82  be  on  interest  to  amount  to  $761.44, 
at  5  per  cent.  ?  Ans.  3  yr.  4  mo.  24  da. 

3.  How  long  will  it  take  any  sum  of  money  to  double  itself  by 
simple  interest  at  3,  4},  6,  7,  and  10  per  cent.  ?  How  long  to 
quadruple  itself?      ^^^  J  To  double  itself  at  3  %,  33J  yr 

^^*  (  To  quadruple  itself  at  3  %.  100  yr. 

4.  In  what  time  will  $9750  produce  $780  interest,  at  2  per 
cent,  a  month  ? 

5.  In  what  time  will  $1000  draw  $1171.353  at  6  per  cent,  com- 
pound interest'/ 

Analysis.  $1171.353-t-1000^$1.171353,  the  amount  of  $1  for  the 
required  time.  From  the  table,  $1,  in  2  years,  will  amount  to  $1.123G ; 
hence  $1.171353— $1.1236=$.047753,  the  interest  which  must  accrue 
on  $1.1230  for  the  fraction  of  a  year;  and  $1.1230  X  .00  =  :i^.0G741G  ; 
$.047753  ~  $.007410  =  .7083  yr.  =  8  mo.  15  da. 

Ans,  2  yr.  8  mo.  15  da. 

6.  In  what  time  will  $333  amount  to  $376.76  at  5  per  cent 
compound  interest,  payable  semi-annually  ? 

7.  In  what  time  will  any  sum  double  itself  at  6  %  compound 
interest?     At  7  %  ?  Ans,  to  last,  10  yr.  2  mo.  26  da. 

DISCOUNT. 

556,  Discount  is  an  abatement  or  aliowance  made  for  the 
payment  of  a  debt  before  it  is  due 

^^7.  The  Present  Worth  of  a  debt,  payable  at  a  future  time 
without  interest,  is  such  a  sum  as,  being  put  at  legal  interest,  will 
amount  to  the  given  debt  when  it  becomes  due. 

1.  What  is  the  present  worth  and  what  the  discount  of  $642.12 
to  be  paid  4  yr.  9  mo.  27  da.  hence,  money  being  worth  7  per 
cent.  ? 


DISCOUNT.  329 

OPERATION.  Analysis.    Since  $1  is  the. 

$1.33775,  Amount  of  Si.  *  present  worth   of  $1.33775 

$642.12  ^  1.33775  =«  §180  for   the  given   time    at   the 

$642.12,  given  sum.  given   rate  of  interest,  the 

480.        present  worth.  present    worth    of   $G42.12 

$162~12^  discount.  ^"^^^  ^^  ^^  many  dollars  as 

$1.33775  is  contained  times 
in  $642.12.  Dividing,  and  we  obtain  $480  for  the  present  worth,  and 
subtracting  this  sum  from  the  given  sum,  we  have  $162.12,  the  dis- 
count.    Hence  the  following 

Rule.  I  Divide  the  given  sum  or  debt  hy  the  amount  of  $1 
for  the  given  rate  and  time;  the  quotient  will  he  the  present  worth 
of  the  debt. 

II.  Subtract  the  present  worth  from  the  given  sum  or  debt;  the 
remainder  will  be  the  discount. 

Notes.  —  1.  The  terms  present  worthy  discount,  and  debt,  are  equivalent  to 
principal^  interest,  and  amount.  Hence,  when  the  time,  rate  per  cent.,  and 
amount  are  given,  the  principal  may  be  found  by  Prob.  II,  (663);  and  the 
interest  by  subtracting  the  principal  from  the  amount. 

2.  When  payments  are  to  be  made  at  different  times  without  interest,  find  the 
present  worth  of  eaoh  payment  {separately.  Their  sum  will  be  the  presewt  worth 
of  the  several  payments,  and  this  sura  subtracted  from  the  sum  of  the  several 
payments  will  leave  the  total  discount. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  present  worth  of  a  debt  of  $385.31^,  to  be  paid 
in  5  mo.  15  da.,  at  6  %  ?  Ans,  $375. 

2.  How  much  should  be  discounted  for  the  present  payment  of 
a  note  for  $429*f  86,  due  in  1  yr.  6  mo.  1  da.,  money  being  worth 
5i  %  ?  Ans,  $32,826. 

3.  Bought  a  farm  for  $2964.12  ready  money,  and  sold  it  again 
for  $3665.20,  payable  in  1  yr.  6  mo.  How  much  would  be  gained 
in  ready  mon^y,  discounting  at  the  rate  of  8  %  ? 

4.  A  man  bought  a  flouring  mill  for  $25000  cash,  or  for  $12000 
payable  in  6  mo.  and  $15000  payable  in  1  yr.  3  mo.  He  accepted 
the  latter  offer;  did  he  gain  or  lose,  and  how  much,  money  being 
worth  to  him  10  per  cent.  ?  Ans.  Gained  $238.10. 

5.  B  bought  a  house  and  lot  April  1,  1860,  for  which  he  waa 
to  pay  $1470  m  the  fourth  day  of  the  following  September,  and 


330  PERCENTAGE. 

$2816.80  Jan  1,  1861.  If  he  could  get  a  discount  of  10  per 
cent,  for  present  payment,  How  much  would  he  gain  by  borrowing 
the  sum  at  7  per  cent.,  and  how  much  must  he  borrow? 

6.  What  is  the  difference  between  the  interest  and  the  discount 
of  $576,  due  1  yr.  4  mo.  hence,  at  6  per  cent.  ? 

7.  A  merchant  holds  two  notes  against  a  customer,  one  for 
$243.16,  due  May  6,  1861,  and  the  other  for  $178.64,  due  Sept. 
25,  1861 ;  how  much  ready  money  would  cancel  both  the  notes 
Oct.  11, 1860,  discounting  at  the  rate  of  7  %  ?   Ans.  $401.29 — . 

8.  A  speculator  bought  120  bales  of  cotton,  each  bale  containing 
488  pounds,  at  9  cents  a  pound,  on  a  credit  of  9  months  for  the 
amount.  He  immediately  sold  the  cotton  for  $6441.60  cash,  and 
paid  the  debt  at  8  %  discount ;  how  much  did  he  gain  ? 

9.  Which  is  the  more  advantageous,  to  buy  flour  at  $6.25  a 
barrel  on  6  months,  or  at  $6.50  a  barrel  on  9  months,  money  being 
worth  8  %  ? 

10.  How  much  may  be  gained  by  hiring  money  at  5  %  to  pay 
a  debt  of  $6400,  due  8  months  hence,  allowing  the  present  worth 
of  this  debt  to  be  reckoned  by  deducting  5  ^o  per  annum  dis- 
count? Ans.  $7.11^. 

BANKING. 

S58.  A  Bank  is  a  corporation  chartered  by  law  for  the  pur- 
pose of  receiving  and  loaning  money,  and  furnishing  a  paper 
circulation. 

^^9.  A  Promissory  Note  is  a  written  or  printed  engagement 
to  pay  a  certain  sum  either  on  demand  or  at  a  specified  time. 

S60.   Bank  Notes,  or  Bank  Bills,  are  the  notes  made  and 

issued  by  banks  to  circulate  as  money.    They  are  payable  in  specie 

at  the  banks.  , 

Note. — A  bank  which  issues  notes  to  circulate  as  money  is  called  a  hanJc  of 
issue ;  one  which  lends  money,  a  bank  of  discount ;  and  one  which  takes  chart^e 
of  money  belonging^  to  other  parties,  a  hank  of  deposit.  Some  banks  perform 
two  and  some  all  of  these  duties. 

^61.  The  Maker  or  Drawer  of  a  note  is  the  person  by  whom 
the  note  is  signed ; 

562.  The  Payee  is  the  person  to  whose  order  the  note  is  made 
payable;  and 


BANKING.  831 

S63.   The  Holder  is  the  owner. 

^04:  A  Negotiable  Note  is  one  which  may  be  bought  and 
sold,  or  negotiated.  It  is  made  payable  to  the  hearer  or  to  the 
order  of  the  payee. 

«i6o«   Indorsing  a  note  by  a  payee  or  holder  is  the  act  of 

writing  his  name  on  its  back. 

Notes. — 1.  If  a  note  is  payable  to  the  bearer,  it  may  be  negotiated  without 
indorsement. 

2.  An  indorsement  makes  the  indorser  liable  for  the  payment  of  a  note,  if  the 
maker  fails  to  pay  it  when  it  is  due. 

3.  A  note  should  contain  the  words  "  value  received,"  and  the  sum  for  which 
it  is  ^iven  should  be  written  out  in  words. 

^06.  The  Face  of  a  note  is  the  sum  made  payable  by  the 
note. 

•5G7.  Days  of  Grace  are  the  three  days  usually  allowed  by 
law  for  the  payment  of  a  note  after  the  expiration  of  the  time 
specified  in  the  note. 

568.    The  Maturity  of  a  note  is  the  expiration  of  the  days 

of  grace  ]  a  note  is  due  at  maturity. 

Note. — No  grace  is  allowed  on  notes  payable  "on  demand,"  without  grace. 
In  some  States  no  grace  is  allowed  on  notes,  and  their  maturity  is  the  wxpira- 
tion  of  the  time  mentioned  in  them. 

5@9*   Notes  may  contain  a  promise  of  interest,  which  will 

be  reckoned  from  the  date  of  the  note,  unless  some  other  time  be 

specified. 

NoTK. — A  note  is  on  interest  from  the  day  it  is  due,  even  though  no  mention 
be  mude  of  interest  in  the  note. 

570.  A  Notary,  or  Notary-Public,  is  an  officer  authorized 
by  law  to  attest  documents  or  writings  of  any  kind  to  make  them 
authentic. 

571.  A  Protest  is  a  formal  declaration  in  writing,  made  by  a 
Notary-Public,  at  the  request  of  the  holder  of  a  note,  notifying 
the  maker  and  the  mdorsers  of  its  non-payment. 

Notes. — 1.  The  fulhire  to  protest  a  note  on  the  tliird  day  of  grace  releases  the  iu- 
dorsers  from  all  obligation  to  pay  it. 

2.  If  the  third  day  of  grace  or  the  maturity  of  a  note  occurs  on  Sunday  or  a  legal 
holiday,  it  must  be  paid  on  the  day  previous. 

«>73«  Bank  Discount  is  an  allowance  made  to  a  bank  for  the 
payment  of  a  note  before  it  becomes  due. 


■ 


3g2  PERCENTAGE. 

^73.  The  Proceeds  of  a  note  is  the  sum  received  for  it  when 
discounted,  and  is  equal  to  the  face  of  the  note  less  the  discount. 

^74:,  The  transaction  of  borrowing  money  at  banks  is  con- 
ducted in  accordance  with  the  following  custom :  The  borrower 
presents  a  note,  either  made  or  indorsed  by  himself,  payable  at  a 
specified  time,  and  receives  for  it  a  sum  equal  to  the  face ;  less 
the  interest  for  the  time  the  note  has  to  run.  The  amount  thus 
withheld  by  the  bank  is  in  consideration  of  advancing  money  on 
the  note  prior  to  its  maturity. 

Notes. — 1.  A  note  for  discount  at  bank  must  be  made  payable  to  the  order 
of  some  person,  by  whom  it  must  be  indorsed. 

2.  The  business  of  buying  or  discounting  notes  is  chiefly  carried  on  by  banks 
and  brokers. 

^73.  The  law  of  custom  at  banks  makes  the  bank  discount 
of  a  note  equal  to  the  simple  interest  at  the  legal  rate,  for  the 
time  specified  in  the  note.  As  the  bank  always  takes  the  interest 
at  the  time  of  discounting  a  note,  bank  discount  is  equal  to  simple 
interest  paid  in  advance.  Thus,  the  true  discount  of  a  note  for 
$153,  which  matures  in  4  months  at  6  %,  is  $153 —  'fsoo  ^ 
$3.00,  and  the  bank  discount  is  $153  x  .02  =  $3.06.  Since  the 
interest  of  $3,  the  true  discount,  for  4  months  is  $3  x  .02  =  $.06, 
we  observe  that  the  bank  discount  of  any  sum  for  a  given  time  is 
greater  than  true  discount,  by  the  interest  on  the  true  discount 
for  the  same  time. 

NoTB.  —  Many  banks  take  only  true  discount. 

CASE   I. 

S7&.  Given,  the  face  of  a  note,  to  find  the  discount 
and  the  proceeds. 

KuLE.  I.  Compute  the  interest  on  tlie  face  of  the  note  for  three 
dey^  more  than  the  specified  time ;   the  residt  will  be  the  discmint, 

II.  Subtract  the  discount  from  the  face  of  the  note;  the  re- 
mainder will  be  the  proceeds. 

NoTKS.  —  1.  When  a  note  is  on  interest,  payable  nt  a  future  specified  time,  the 
omomit  is  the  face  of  the  note,  or  the  sura  made  payable,  and  must  be  made  the 
b;i.«jis  01  discount. 

2.  To  indicate  the  maturity  of  a  note  or  draft,  a  vertical  line  (  |  )  is  used,  with 
the  day  at  which  the  note  is  nominally  due  on  the  left,  and  the  date  of  maturity 
(Ml  tk«  wght;  thug,  Jan.  "^  |  jq. 


BANKING.  333 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  bank  discount,  and  what  are  the  proceeds  of  a 
note  for  §1487  due  in  30  days  at  6  per  cent.  ? 

Ans.  Discount,  ^8.18;  Proceeds,  $1478.82. 

2.  What  are  the  proceeds  of  a  note  for  $884.50  at  90  days,  if 
discounted  at  the  New  York  Bank? 

3.  Wishing  to  borrow  $1000  of  a  Southern  bank  that  is  dis- 
counting paper  at  8  per  cent.,  I  give  my  note  for  $975,  payable 
in  60  days ;  how  much  more  will  make  up  the  required  amount  ? 

4.  A  man  sold  his  farm  containing  195  A.  2  K.  25  P.  for  $27.59 
an  acre,  and  took  a  note  payable  in  4  mo.  15  da.  at  7  %  interest. 

^  Wishing  the  money  for  immediate  use,  he  got  the  note  discounted 
I  at  a  bank;  how  much  did  he  receive?  Ans.  $5236.169. 

5.  Find  the  day  of  maturity,  the  term  of  discount,  and  the  pro- 
ceeds of  the  following  notes : 

$1962^.  Detroit,  July  26,  1860. 

Four  months  after  date  I  promise  to  pay  to  the  order  of  James 
Gillis  one  thousand  nine  hundred  sixty-two  and  j^j^^^  dollars  at  the 
Exchange  Bank,  for  value  received.  John  Demar^st. 

Discounted  Aug.  26,  at  7%. 

Ans.  Due  Nov.  ^^  |  09;  term  of  discount  95  days;  preoec^, 
$1926.20. 


$1066yY^.  Baltimore,  April  19,  1859. 

6.  Ninety  days  after  date  we  promise  to  pay  to  the  order  of 
King  &  Dodge  one  thousand  sixty-six  and  -^j^j^  dollars  at  the  Citi* 
zens'  Bank,  for  value  received.  Case  &  Sons. 

Discounted  May  8,  at  6  %. 

Ans.  Due  July  » »  |  , , ;  term  of  discount,  74  da. ;  proceeds, 
$1058.59. 

$784^.  .  Mobile,  June  20,  1861. 

7.  Two  months  after  date  for  value  received  I  promise  to  pay 
George  Thatcher  or  order  seven  hundred  eighty-four  and  -^^^^  dol* 
lars  at  the  Traders'  Bank.  Wm.  Hamilton. 

Discounted  July  5,  at  8  ^. 


■ 


834  PERCENTAGfB. 


Sl845^<\j.  Chicago,  Jan.  31,  1862. 

8.  One  inontli  after  date  we  jointly  and  severally  agree  to  pay 
to  W.  H.  Willis,  or  order,  one  thousand  eight  hundred  forty-five 
and  -f^Q  dollars  at  the  Marine  Bank. 

Payson  &  Williams. 

Discounted  Jan.  31,  at  2  %  a  month. 

Ans.  Due  Feb.  28  |  March  3;  term  of  discount,  31  da.;  pro-, 
ceeds,  $1807.36. 

9.  What  is  the  difierence  between  the  true  and  the  bank  dis- 
count of  $950,  for  3  months  at  7  per  cent.  ?  Ans.  $.29. 

10.  What  is  the  difference  between  the  true  and  the  bank  dis- 
count of  $1375.50,  for  60  days  at  6  per  cent.  ? 

CASE   II. 

577.   Given,  the  proceeds  of  a  note,  to  find  the  face. 
*1.  For  what  sum  must  I  draw  my  note  at  4  months,  interest 
6  %,  that  the  proceeds  when  discounted  in  bank  shall  be  $750  ? 

OPERATION.  Analysis.       We 

$1.0000  first  obtain  the  pro- 

.0205,  disc't  on  $1  for  4  mo.  3  da.         ceeds  of  $1  by  the 

$~^795,  proceeds  of  $1.  ^^^* ^^^^'  *^^^'  '^"^^ 

$750  --  .9795  =  $765,696,  Ans.  ^-^^^^  is  *h^  P^^ 

ceeds  of  $1,  $750  is 

the  proceeds  of  as  many  dollars  as  $.9795  is  contained  times  in  $750. 

Dividing,  we  obtain  the  required  result.     Hence  the 

Rule.  Divide  the  proceeds  hy  the  proceeds  of  $iybr  the  time 
and  rate  mentioned  ',   the  quotient  will  he  the  face  of  the  note, 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  face  of  a  note  at  60  days,  the  proceeds  of  which, 
when  discounted  at  bank  at  6  %,  are  $1275?    Ans,  $1288.53. 

2.  If  a  merchant  wishes  to  draw  $5000  at  bank,  for  what  sum 
must  he  give  his  note  at  90  days,  discounting  at  6  per  cent.  ? 

Ans.  $5078.72. 

3.  The  avails  of  a  note  having  3  months  to  run,  discounted  at 
a  bank  at  7  %,  were  $276.84;  what  was  the  face  of  the  note? 


BANKING.  335 

4.  James  T.  Fisher  buys  a  bill  of  merchandise  in  New  York  at 
cash  price,  to  the  amount  of  $1486.90,  and  gives  in  payment  his 
note  at  4  months  at  7A  % ;  what  must  be  the  face  of  the  note  ? 

5.  Find  the  face  of  a  6  mo.  note,  the  proceeds  of  which,  dis- 
counted at  2  %  a  month,  are  $496.  Ana.  $564.92. 

6.  For  what  sum  must  a  note  be  drawn  at  30  days,  to  net 
$1200  when  discounted  at  5  %  ? 

7.  Owing  a  man  $575,  I  give  him  a  60  day  note ;  what  should 
be  the  face  of  the  note,  to  pay  him  the  exact  debt,  if  discounted 
at  1|  %  a  jmonth?  Ans.  $593.70. 

8.  What  must  be  the  face  of  a  note  which,  when  discounted  at 
a  broker's  for  110  days  at  1  5^  a  month,  shall  give  as  its  proceeds 
$187.50? 

CASE   III. 

578.  Given,  the  rate  of  bank  discount,  to  find  the 
corresponding  rate  of  interest. 

I.  A  broker  discounts  30  day  notes  at  1^  ^  a  month;  what 
rate  of  interest  does  his  money  earn  him  ? 

OPERATION.  Analysis.  If  we  assume 

30  day  notes  »=  33  days'  time.  $100  as  the  face  of   the 

$100,         base.  note,  the   discount  for  33 

1.65,  discount  for  33  days.  days  at  IJ^  a  month  will 

$98.35,  proceeds.  ^^  ^^'^^  ^"^  ^^^  proceeds 

$1.65 -f-. 090154 J  =18AV^%,^7is.    $98.35.      We    then    have 

$98.35  principal,  $1.65  in- 
terest, and  33  days  time,  to  find  the  rate  per  cent,  per  annum,  which 
we  do  by  (554).     Hence  the 

Rule.  I.  Find  the  discount  and  the  proceeds  of  $1  or  $100 
for  the  time  the  note  has  to  run. 

II.  Divide  the  discount  hy  the  interest  of  the  proceeds  at  1  per 
cent,  for  the  same  time. 

EXAMPLES    FOR   PRACTICE. 

1.  "What  rate  of  interest  is  paid,  when  a  note  payable  in  30 
days  is  discounted  at  6  per  cent.?  Ans.  ^H^  %• 


336  PERCENTAGE. 

2.  A  note  payable  in  2  months  is  discounted  at  2  %  a  month; 
"what  rate  of  interest  is  paid?  Ans.  2b-^^^^  %. 

3.  When  a  note  ^payable  in  90  days  is  discounted  at  IJ  %  a 
month,  what  rate  of  interest  is  paid?  Ans,  18yf  Jf  %• 

4.  What  rate  of  interest  corresponds  to  5,  6,  7,  10,  12  %  dis- 
count on  a  note  running  10  months  without  grace  ? 

5.  What  rate  of  interest  does  a  man  pay  who  has  a  60  day 
note  discounted  at  |,  1,  2,  2^,  3  %  a  month  ? 

CASE    IV. 

579.   Given,  tke  rate  of  interest,  to  find  the  corres- 
ponding rate  of  bank  discount. 

1    A  broker  buys  60  day  notes  at  such  a  discount  that  his 

money  earns  him  2  ^  a  month;  what  is  his  rate  ^  of  discount? 

OPERATION.  Analysis.     If  we  assume 

60  da.  -f  3  da.  =  63  da.  ^1^^  a,s  the  proceeds  of  a 

$100  base.  note,  the  interest  for  63  days 

4.20,  interest  for  63  da.  at  24  per  cent,  will  be  $4.20, 

il0r20,  amount       "       ^^  ^"^  *^^  ^"'^""^  ^^  ^^^^  ^^ 

$4.20  -f-  .18235  =  23^^^-  %,  Ans.    the  note  will  be  $104.20.  We 

then  have  $104.20  the  prin- 
cipal, $4.20  the  interest,  and  63  days  the  time,  to  find  the  rate  per 
cent.,  which  we  do  by  (549)  as  in  the  last  case.     He^oe  the 

Rule.     I.   Find  the  interest  and  the  amount  o/%\  or  §100 /or 
the  time  the  note  has  to  run, 

II.  Divide  the  interest  hy  the  interest  on  the  amount  at  1  per 
cent,  for  the  same  time. 

EXAMPLES   FOR   PRACTICE. 

1.  W^hat  rates  of  bank  discount  on  30  day  notes  correspond  to 
5,  6,  7,  10  per  cent,  interest? 

2.  At  what  rate  should  a  3  months^  note  be  diseounted  to  pro- 
duce 8  %  interest?  Ans.  Vjlf ?   %. 

3.  At  what  rates  should  60  day  notes  be  discounted  to  pay  to  a 
•broker  1,  li,  2,  2}  %  a  month? 

4.  At  what  rate  must  a  note  payable  18  months  hence,  without 
grace,  be  discounted  to  produce  7  %  interest?     Ans.  6/^*^  %• 


EXCHANGE.  837 

EXCHANGE. 

580.  Exchange  is  a  method  of  remitting  money  from  one 
place  to  another,  or  of  making  payments  by  written  orders. 

581.  A  Bill  of  Exchange  is  a  written  request  or  order  upon 
one  person  to  pay  a  certain  sum  to  amother  person,  or  to  his  order, 
at  a  specified  time. 

58S.  A  Sight  Draft  or  Bill  is  one  requiring  payment  to  be 
made  "  at  sight,''  which  means,  at  the  time  of  its  presentation  to 
the  person  ordered  to  pay.  In  other  bills,  the  time  specified  is 
usually  a  certain  number  of  days  ''  after  sight/' 

There  are  always  three  partiQ3  to  a  transaction  in  exchange,  and 
usually  four : 

I58S.  The  Drawer  or  Maker  Is  the  person  who  signs  the 
order  or  bill ; 

584.  The  Drawee  is  the  person  to  whom  ,the  order  is  ad- 
dressed ; 

585.  The  Payee  is  the  person  to  whom  the  money  is  ordered 
to  be  paid  ]  and 

586.  The  Buyer  or  Remitter  is  the  person  who  purchases 
the  bill.  He  may  be  himself  the  payee,  or  the  bill  may  be  drawn 
in  favor  of  any  other  person. 

587.  The  Indorsement  of  a  bill  is  the  writing  upon  its  back, 
by  which  the  payee  relinquishes  his  title,  and  transfers  the  pay- 
ment to  another.  The  payee  may  indorse  in  blank  by  writing  his 
name  only,  which  makes  the  bill  payable  to  the  hearer ,  and  con- 
sequently transferable  like  a  bank  note ;  or  he  may  accompany  his 
signature  by  a  special  order  to  pay  to  another  person,  who  in  his 
turn  may  transfer  the  title  in  like  manner.  Indorscrs  become  sep- 
arately responsible  for  the  amount  of  the  bill,  in  case  the  drawee 
fails  to  make  payment.  A  bill  made  payable  to  the  hearer  is 
transferable  without  indorsement. 

588.  The  Acceptance  of  a  bill  is  the  promise  which  the 
drawee  makes  when  the  bill  is  presented  to  him  to  pay  it  at  ma- 
turity; this  obligation  is  usually  acknowledged  by  writing* thp 
word  "  Accepted,"  with  his  signature,  across  the  face  of  the  bill* 

29  w 


838  PERCENTAGE. 

Notes. — I.  In  this  country,  and  in  Great  Britain,  three  days  of  grace  are  al- 
lowed for  the  payment  of  a  bill  of  exchange,  after  the  time  specitied  has  expired. 
In  regard  to  grace  on  siglit  hills,  however,  custom  is  variable  ;  in  New  York. 
Penn^ivlvania,  Virginia,  and  some  other  States,  no  grace  is  allowed  on  sight  bills, 

2.  ^V'hen  a  bill  is  protested  for  non-acceptance,  the  drawer  is  obligated  to  pay 
it  immediately,  even  though  the  specified  time  has  not  expired. 

Exchange  is  of  two  kinds  —  Domestic  and  Foreign. 

58l>.    Domestic  or  Inland  Exchange  relates  to  remittances 

made  between  different  places  of  the  same  country. 

Note. — An  Inland  Bill  of  Exchange  is  commonly  called  a  Draft. 

•5@0o  Foreign  Exchange  relates  to  remittances  made  between 
different  countries. 

*5IJ1.  A  Set  of  Exchange  consists  of  three  copies  of  the  same 
bill,  made  in  foreign  exchanges,  and  sent  by  different  conveyances 
to  provide  against  miscarriage;  when  one  has  been  paid,  the  others 
are  void. 

^1^^.  The  Face  of  a  bill  of  exchange  is  the  sum  ordered  to 
be  paid ;  it  is  usually  expressed  in  the  currency  of  the  place  on 
which  the  draft  is  made. 

593.  The  Par  of  Exchange  is  the  estimated  value  of  the 
coins  of  one  country  as  compared  with  those  of  another,  and  is 
either  intrindc  or  commercial. 

t594.  The  Intrinsic  Par  of  Exchange  is  the  comparative 
value  of  the  coins  of  different  countries,  as  determined  by  their 
weight  and  purity. 

Q^^»   The  Commercial  Par  of  Exchange  is  the  comparative 

value  of  the  coins  of  different  countries,  as  determined  by  their 

nominal  or  market  price. 

Note. — The  intrinsic  par  is  always  the  same  while  the  coins  remain  un- 
•  changed;  but  the  commercial  par,  being  determined  by  commercial  usage,  is 
fluctuating. 

^9G.  The  Conrse  of  Exchange  is  the  current  price  paid  in 
one  place  for  bills  of  exchange  on  another  place.  This  price 
varies,  according  to  the  relative  conditions  of  trade  and  commercial 
credit  at  the  two  places  between  which  exchange  is  made.  Thus, 
if  Boston  is  largely  indebted  to  Paris,  bills  of  exchange  on  Paris 
will  bear  a  high  price  in  Boston. 

When  the  course  of  exchange  between  two  places  is  unfavor- 


EXCHANGE.  839 

able  to  drawing  or  rcinitting,  the  disadvantage  is  sometimes 
avoided,  by  means  of  a  circuitous  exchange  on  intermediate  places 
between  which  the  course  is  favorable. 

DIRECT    EXCHANGE. 

^97.  Direct  Exchange  is  confined  to  the  two  places  between 
which  the  money  is  to  be  remitted. 

j     598.    There  are  always  two  methods  of  transmitting  money 
'  between  two  places.     Thus,  if  A  is  to  receive  money  from  B, 

1st.  A  may  draw  on  B,  and  sell  the  draft; 

2d.  B  may  remit  a  draft,  made  in  favor  of  A. 

Note.  —  One  person  is  said  to  draw  on  another  person,  when  he  is  the  maker 
of  a  draft  addressed  to  that  person. 

CASE   I. 

599.   To  compute  domestic  exchange. 

The  course  of  exchange  for  inland  bills,  or  drafts,  is  always  ex- 
pressed by  the  rate  of  premium  or  discount.  Drafts  on  time, 
however,  are  subject  to  hank  discount ,  like  notes  of  hand,  for  the 
term  of  credit  given.  Hence,  their  cost  is  affected  by  both  the 
course  of  exchange  and  the  discount /or  time. 

1,  What  will  be  the  cost  of  the  following  draft,  exchange  on 
Boston  being  in  Pittsburgh  at  2}  ^  premium  ? 

$600.  Pittsburgh,  June  12,  18G0. 

Sixty  days  after  sight,  pay  to  William  Barnard,  or  order,  six 
hundred  dollars,  value  received,  and  charge  the  same  to  our 
account. 

To  the  Suffolk  Bank,  Boston.  Thomas  Bauer  &  Co. 

OPERATION. 

$1  4-  $.0225  ==  $1.0225,  course  of  exchange. 

.0105,  bank  discount  of  $1,  (63  da.) 

$1,012,    cost  of  exchange  for  $1. 
S600  X  1.012  =  $607.20,^728. 
NALTSis.     From   $1.0225,  the   course  of  exchange,  we   subtract 
$.0105,  the  bank  discount  of  $1  for  the  specified  time,  and  obtain 
$1,012,  the  cost  of  exchange  for  $1 ;  then  $G00  X  1.012  =  $G07.20,  the 
eost  of  exchange  for  $600. 


340  PERCENTAGE. 

2.  A  commission  merchant  in  Detroit  wishes  to  remit  to  his 
employer  in  St.  Louis,  $512.36  by  draft  at  60  days ;  what  is  the 
face  of  the  draft  which  he  can  purchase  with  this  sum,  exchange 
being  at  2  i  %  discount  ? 

OPERATION. 

$1  —  S.025  =  ?.975,      course  of  exchange. 
.01225,  discount  of  §1. 

$.96275,  cost  of  exchange  for  $1. 
?512.36  ~  .96275  =  §532.18+,  Ans. 

Analysis.  From  $.975,  the  course  of  exchange,  we  subtract 
$.01225,  the  bank  discount  of  $1  for  the  specified  time,  at  the  legal 
rate  in  Detroit,  and  obtain  $.96275,  the  cost  of  exchange  for  $1 ;  and 
the  face  of  the  draft  that  will  cost  $512.36,  will  be  as  many  dollars  as 
$.90275  is  contained  times  in  512.36,  which  is  532.18+,  times. 
Hence  we  have  the  following 

Rule.     I.  To  find  the  cost  of  a  draft,  the  face  being  given 

Multiply  the  face  cf  the  draft  hy  the  cost  of  exchange  for  $1. 

II.  To  find  the  face  of  a  draft,  the  cost  Doing  given.  —  Divide 

the  given  cost  hy  the  cost  of  exchange  for  $1. 

Note.  —  The  cost  of  exchange  for  $i  may  always  be  found,  by  subtracting 
from  the  course  of  exchange  the  bank  discount  (at  the  legal  rate  where  the  draft 
is  made),  for  the  specified  time.  Foi  sight  drafts,  the  course  of  exchange  is  tho 
cost  of  $1. 

EXAMPLES   FOR   PRACTICE. 

1.  What  must  be  paid  in  New  York  for  a  draft  on  Boston,  at 
80  days,  for  §5400,  exchange  being  at  J  %  premium  ? 

Ans.  $5392.35. 

2.  What  is  the  cost  of  sight  exchange  on  New  Orleans,  for 
$3000,  at  3}  %  discount? 

3.  What  must  be  paid  in  Philadelphia  for  a  draft  on  St.  Paul 
drawn  at  90  days,  for  $4800,  the  course  of  exchange  being 
lOlf  %  ?  Ans.  $4791.60. 

4..  A  sight  draft  was  purchased  for  $550.62,  exchange  being  at 
a  premium  of  3^  ^  ;   what  was  the  face  ? 

5.  An  agent  in  Syracuse,  N.  Y.,  having  $1324.74  due  his  em- 
ployer, is  instructed  to.  remit  the  same  by  a  draft  drawn  at  30 
days;  what  will  be  the  face  of  the  draft,  exchange  being  at  If  % 
premium?  Ans.  $1310.22—. 


EXCHANGE.  841 

6.  My  agent  in  Charleston,  S.  C,  sells  a  house  and  lot  for 
$7500,  on  commission  of  IJ  %,  and  remits  to  me  the  proceeds  in 
a  draft  purchased  at  J  %  premium ;  what  sum  do  I  receive  from 
the  sale  of  my  property  ? 

7.  A  man  in  Hartford,  Conn., has  $4800  due  him  in  Baltimore; 
how  much  more  will  he  realize  by  making  a  draft  for  this  sum  on 
Baltimore  and  selling  it  at  J  %  discount,  than  by  having  a  draft 
on  Hartford  remitted  to  him,  purchased  in  Baltimore  for  this  sum 
at  I  %  premium?  Ans,  $11.73  +  . 

8.  The  Merchants^  Bank  of  New  York  having  declared  a  divid- 
end of  6}  %,  a  stockholder  in  Cincinnati  drew  on  the  bank  for  the 
sum  due  him,  and  sold  the  draft  at  a  premium  of  If  %,  thus  real- 
izing $508.75  from  his  dividend;  how  many  shares  did  he  own  ? 

9.  Sight  exchange  on  New  Orleans  for  $5000  cost  $5075; 
what  was  the  course  of  exchange  ?  Ans.   IJ  %  premium. 

10.  A  man  in  Buffalo  purchased  a  draft  on  St.  Paul,  Minn., 
for  $5320,  drawn  at  60  days,  paying  $5141.78;  what  was  the 
course  of  exchange  ?  -4»s.  2i  %  discount. 

CASE   IT. 

®00.   To  compute  foreign  exchange. 
001.    The  following  standards  of  the  decimal  currency  of  the 
United  States  were  established  April  2,  1792. 

Coins.  Weight.  Fineness. 

Gold  eagle, 270  grains,     91Gf  thousandths. 

Silver  dollar, 416       '' 

Copper  cent, 264      " 

In  1834,  the  eagle  was  reduced  in  weight  to  258  grains,  and  in  1837 
its  fineness  was  fixed  at  900  thousandths  pure,  which  is  likewise  the 
present  standard  of  purity  for  all  the  U.  S.  gold  and  silver  coins.  In 
1837,  also,  the  silver  dollar  was  reduced  in  weight  to  412.5  grains. 
In  1853,  the  silver  half  dollar  was  reduced  in  weight  to  192  grains, 
and  the  smaller  silver  coins  proportionally. 

NoTn.  —  The  object  of  the  change  in  the  silver  coinage  of  the  United  States, 
made  in  1853,  was  to  prevent  its  exportation  bj  raiding  the  nominal  value  of 
silver  above  its  foreign  market  value. 

®0^»   The    intrinsic  par   of  exchange  between    the    United 
States  and  different  countries,  is  given  in  the  following 
29* 


342  PERCENTAGE. 

TABLE   OP   FOREIGN    COINS   AND    MONEY. 


Crown,  Baden 

"       Bavaria 

"       England 

"       France 

"       Geneva 

"       Portugal 

"       Tuscany 

"      Wurtemberg 

•   "       Zurich 

Dollar,  Argentine  Republic 

"      Bolivia 

"      Chili 

"      Columbia 

"      Mexico 

"      Norway 

"      Peru 

''      Spain 

"      Sweden 

Doubloon,  Bolivia 

"  Columbia  (Bogota)... 

"  "          (Popayau). 

«  Chili  (since  1835) 

"  "    (before  1835)..  .. 

•*  La  Plata 

'*  Mexico  (average) 

«  Peru  (Cuzco) 

"  "     (Lima) 

«  Spain 

Drachma,  Greece 

Ducat,  Austria 

"      Bavaria..... 

"      Cologne 

*'     Hamburg 

'•      Hungary 

"      Netherlands., 

"      Saxony 

"      Sweden 

"      Wurtemberg 

Florin,  Austria 

"      Bavaria 

"      Hanover 

"      Italy 


Gold. 
Silver. 


Gold. 


Silver. 
Gold. 


Silv 


£>euouiiiations. 


8  reals. 
8       " 
100  cents. 
8  reals. 
8       " 
6  marks. 
8  reals. 
10     "     (old). 
6  marks. 


4  gilders 
12  marks. 

6C  kreutzers. 
60  « 

60  groshen. 
12  soldi. 


i 

1.077 

1.157 

1.072 

1.151 

1.100 

1.181 

1.100 

1.181 

.960 

1.031 

5.813 

1.050 

1.128 

1.070 

1.148 

.960 

1.031 

1.016 

1.091 

1.011 

1.086 

1.011 

1.080 

1.022 

1.09S 

1.005 

1.079 

1.051 

1.129 

1.005 

1.079 

1.003 

1.077 

1.059 

1.136 

15.580 

15.617 

15.390 

15.060 

15.570 

14.060 

15.534 

15.534 

15.551 

15  570 

166 

.177 

2.278 

2.274 

2.250 

2.257 

2.2S1 

2.269 

2.264 

2.267 

2Ji36 

.485 

.521 

.395 

.425 

.547 

.587 

.181 

.194 

EXCHANGE.  343 

TABLE    OP   FOREIGIT    COINS   AND   MONEY  — Continued. 


Florin,  Mecklenburg 

"      Prussia  and  Poland. 

"      Tuscany 

Franc,  Belgium 

'•      France 

Frederick  d'or,  Denmark 

Gilder,  Baden 

"      Darmstadt 

"      Demerara 

"      Frankfort 

"      Netherlands 

"      Wurtemberg 

Ghersh,  Tripoli 

Guinea,  England 

Lira,  Lombardy 

"    Leghorn 

«    Milan 

"    Yenice 

Livre,  Genoa 

"      Leghorn 

"      Switzerland 

Mark  banco,  Hamburg 

"      current,      "        

Milree,  Azores 

"        Brazil...' 

"       Madeira 

«        Portugal 

Mohur,  Ilindostan 

Ounce,  Naples 

Pagoda,  Madras , 

Piaster,  Tunis 

"        Turkey 

Pi.«tareen,  Spain 

Pistole,  Spain 

Pound,  British  ProTinces.... 
Ileal,  plate,  Spain 

"      vellon,     "     

"      Egypt 

Rix  dollar,  Austria 

"        '•      Batavia , 

«        «      Denmark 


Silver. 


Gold. 
Silver. 


Gold. 
Silver. 


Gold. 


Silver. 


Lower 
Denoiniuations. 


30  groshen. 

12  soldi. 
100  centimes. 
100  " 

60  kreutzers. 
60        " 
'    20  stivers. 
60  kreutzers. 

20  stivers. 
60  kreutzers. 

100  paras. 

21  shillings. 
20  soldi. 

20      « 
20      " 

100  centimes. 
20  soldi. 
20      " 
100  centessini. 
16  skilliugs 
10        » 
loco  reis. 
1000      " 
1000      " 
1000       " 
16  rupees. 
3  ducats, 
42  fan  am  s. 
16  carobas. 
100  aspers. 
4  reals  vellon. 


20  shillings. 
34  marvedis. 
34  " 

20  piasters. 
120  kreutzers. 
4S  stivers. 
9u  ^killings. 


1?° 

i'-iS 

5.^?. 

ML 

pm 

.541 

.571 

.227 

.244 

.202 

.281 

.ISO 

.200 

.180 

.200 

3.932 

.397 

.420 

.397 

.426 

.202 

.282 

.397 

.426 

.400 

.436 

.395 

.423 

.105 

.112 

5:059 

.162 

.173 

.162 

.173 

.162 

.173 

.162 

.173 

.186 

.L98 

.102 

.173 

.273 

.292 

.350 

.375 

.2S5 

.3-J5 

.830 

.800 

.830 

.890 

1.000 

1074 

1.120 

1.203 

7.109 

2.485 

1.840 

.124 

.133 

.020 
Xj7 

.028 
.211 

3.904 

4.016 

.097 

.104 

.0{8 

.051 

.DCS 

l.OiO 

.971 

1.043 

,7S2 

.840 

1 

l.Oil 

..,.0 

844  PERCENTAGE. 

TABLE   OF  FOREIGN   COINS   AND   MONEY  —  Continued. 


Rigsbank  dollar 

Rix  dollar,  Norway 

Rouble,  Russia 

Rupee,  India 

Ruspone,  Tuscany 

Sequin,  Tuscany 

Scudo,  Milan 

"     Naples 

"     Rome 

«     Sicily : 

Sovereign,  Great  Britain. 
Thaler,  Brunswick 

"      Hanover 

"      Hesse-Cassel 

"      Prussia 

"      Saxony 

"      Bremen 

Tale,  China 

"    Japan , 

Tomaun,  Persia 

Utchlik,  Tripoli 

Yirmilik,  Turkey , 


Silver. 


Gold. 
Silver. 


Gold. 
Silver. 


Gold. 
Silver. 
Gold. 


Lower 
Denominations. 

-1: 

«  ~  bo 

:i5|i 

48  skillings. 

.526 

.565 

96        « 

1.051 

1.129 

100  copecks. 

.754 

.806 

16  annas. 

6.925 
2.301 

.445 

.477 

117  soldi. 

.973 

1.045 

12  carlini. 

.950 

1.021 

1.006 

1.080 

12  tari. 

.985 

1.058 

20  Bhillings. 

4.861 

30  groschen. 

.692 

.743 

30        « 

.694 

.735 

30        « 

.687 

.738 

SO        « 

.692 

.743 

30        « 

.694 

.735 

72  grotes. 

.788 

.846 

10  mace,  100  -» 
candarines.  S 

1.480 

1.590 

10  mace,  100  \ 
candarines.  i 

.760 

.800 

100  mamvodis. 

2.233 

120  paras. 

.149 

.160 

20  piasters. 

.877 

Notes. — 1.  The  standard  value  of  gold  a.s  compared  with  silver  in  the  United 
States,  is  as  15.407  to  1  in  the  coinage  of  1792,  as  15.988  to  1  in  the  coinage  of 
.  1837,  nnd  as  14.922  to  1  in  the  coinage  of  1853. 

2.  The  relative  values  of  gold  and  silver  differ  in  the  coinage  of  different  coun- 
tries. In  England  the  ratio  is  14.288  to  1  j  in  France  it  is  15.5  to  1  ;  in  Ham- 
burg it  is  15  to  1. 

3.  In  the  present  gold  coinage  of  the  United  States,  a  Troy  ounce  of  pure  gold 
is  equal  to  $20,672,  and  of  standard  gold  to  $18,605.  In  the  present  silver 
coinage  of  the  United  States,  a  Troy  ounce  of  pure  silver  is  equal  to  $1,388,  and 
of  standard  silver  to  $1.25. 

603.  It  will  be  seen  by  the  table  that  the  par  of  exchange 
between  the  United  States  and  Great  Britain  is  £1  =  §4.861. 
Previous  to  the  changes  in  the  U.  S.  coinage^  made  in  1834  and 
in  1837,  the  par  of  exchange  was  £1  =  S4.44|,  or  £9  =  §40, 
which  is  called  the  old  par  of  excfiange.     By  the  new  par  of  ex- 


EXCHANGE.  345 

change,  sterling  money  is  worth  about  9|  %  more  than  by  the 
old  par. 

@04:.  The  course  of  exchange  on  England  is  usually  given 
with  reference  to  the  old  par  of  exchange.  Hence,  when  sterling 
money  is  really  at  par ^  according  to  present  standards,  it  is  quoted 
in  the  market  at  9|  %  premium. 

60S.  The  course  of  exchange  between  different  countries  may 
be  expressed  either  by  the  rate  per  cent,  above  or  below  par,  or 
by  giving  the  sum  of  money  in  one  country  which  is  equal  to  a 
certain  sum  in  another  country.  In  the  latter  case,  the  exchange 
requires  simply  a  reduction  of  currencies ;  in  the  former,  it  requires 
both  a  reduction  of  currencies  and  a  computation  of  percentage. 

1.  What  will  be  the  cost  in  Boston  of  the  following  bill  of  ex' 
change  on  Liverpool,  at  9 J  %  premium? 


^^^2-  Boston,  June  16,  1860. 

At  sight  of  this  First  of  Exchange  (Second  and  Third  of 
same  tenor  and  date  unpaid)  pay  to  the  order  of  J.  Simmons, 
Boston,  Four  Hundred  Thirty-two  Pounds,  value  received,  and 
charge  the  same  to  account  of 

James  Lowell  &  Co. 
To  Richard  Evans  k  Son,  ) 
Liverpool y  England.  j 

Analysis.      Since 

OPERATION.  exchange   on   Liver- 

£9  =  §40  X  1 .095,  course  of  exchange,      pool  is  at  9 J  %  pre- 

$40  X  1.095  mium,   £9  will  co-^t 

^^  = 9 '  ''''  '^  ^^'  $40  X  1.095,  (603) : 

Aoo      $40x1.095       ^oino  4A      A  and  £1  will  therefore 

432  X ^^ =§2102.40,^....  ^^^    ^    ^,,5 

cost  — g . 

Multiplying  the  face  of  the  bill,  £432,  by  the  cost  of  exchange  of  £1, 
we  obtain  $2102.40,  the  required  cost  of  the  bill. 

2.  What  is  the  face  of  a  bill  on  London,  that  may  be  purchased 
in  New  York  for  $2768.70,  exchange  being  at  10  %  premium  in 
favor  of  London  ? 


346  PERCENTAGE. 

OPERATION. 

£9  =  $40  X  1.10,  course  of  exchange, 
£1  = — '- — J  cost  of  £1, 

$2768.70  -^        \  '      =£bQQ  6s.  6d.,  Ans. 
y 

40  X  1.10 
Analysis.     We  divide  $2768.70,  the  given  cost,  by g— ^ — ,  the 

cost  of  exchange  for  £1,  and  obtain  £566  6s.  6d.,  the  face. 

3.  What  cost,  in  Hamburg,  a  bill  on  New  Orleans  for  $4500,  the 

course  bf  exchange  being  1  mark  =  $.365  ? 

OPERATION.  Analysis.    Since 

^l=-\%%^  of  a  maik,  cost  of  a  unit.  exchange  for  §1  will 

$4500  xVg00^12328  marks  12  skillin^s.     ^^«*    ^^     Hamburg 

iJ^V  of  a  mark,  a  bill 

for  $4500  will  cost  4500  X'||3«=  12328  marks  12  skillings. 
G06.    From  these  illustrations  we  derive  the  following 
Rule.     I.  To  find  the  cost  of  a  bill,  the  face  being  given.  — 

Multi2)h/  the  face  hy  the  co$t  of  a  unit  of  the  currency  in  which  the 

hill  is  expressed. 

II.  To  find  the  face  of  a  bill,  the  cost  being  given. — Divide  the 

given  cost  by  the  cost  of  a  unit  of  the  currency  in  which  the  hill  is 

to  he  expressed, 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  cost  in  Portland  of  a  bill  on  Manchester,  Ene:., 
f^  £325  3s.  9d.,  at  Of  %  premium?  Ans,  $1586.19  +  . 

2.  What  must  be  paid  in  Charleston  for  a  bill  of  exchange  on 
Paris  for  6000  francs,  at  18f  cents  per  franc? 

3.  What  cost  in  Bostc^^  bill  on  St.  Petersburg  for  3000  roubles 
at  IJ  %  premium,  the  |^ar  of  exchtinge  being  $.754  for  1  rouble? 

4.  What  will  be  the  cost  in  Naples  of  a  bill  of  exchange  on 
New  York  fdi-  $831.12,  at  the  rate  of  $.96  for  1  scudo? 

Ans.  865  scudi  9  carlini. 

5.  A  draft  on  Philadelphia  cost  £125  in  Birmingham,  Eng., 
exchange  being  at  8  %  premium  for  sterling;  required  the  face 
of  the  draft. 


EXCHANGE.  34:7 

6.  An  agent  in  Boston,  having  $7530.30  due  liis  emploj^er  in 
England,  is  directed  to  remit  by  a  bill  on  Liverpool )  what  is  the 
face  of  the  bill  which  he  can  purchase  for  this  money,  exchange 
being  at  11  %  premium?  An^^  £1527  12s.  Gj^d. 

7.  A  merchant  in  Cincinnati  has  9087  gilders  10  stivers  due 
him  in  Amsterdam^  and  requests  the  remittance  by  draft;  what 
sum  will  he  receive,  exchange  on  U.  S.  being  in  Amsterdam  at  2  J 
gilders  for  ?1  ? 

8.  A  trader  in  London  wishes  to  invest  £2500  in  merchandise 
in  Lisbon ;  if  he  remits  to  his  correspondent  at  Lisbon  a  bill  pur- 
chased for  this  sum,  at  the  rate  of  64. 5d.  sterling  per  milree,  what 
sum  in  the  currency  of  Portugal  will  the  agent  receive  ? 

Am.  9302  milreeo  325 Jf  reis. 

9.  A  draft  on  Dublin  for  £360  cost  $1736;  what  was  the 
course  of  exchange?  A'lu.  8 J  %  premium. 

10.*  A  merchant  in  Baltimore,  having  receiv(?d  an  importation 
of  Madeira  wine  invoiced  at  1500  milrees,  allows  his  correspondent 
in  Madeira  to  draw  on  him  for  the  sum  necessary  to  cover  the 
cost,  exchange  on  the  United  States  being  in  Madeira  930  rcis  = 
§1 :  how  much  would  the  merchant  have  saved,  by  remitting  a 
draft  on  Madeira,  purchased  at  $1,065  per  1  milree  ? 

Am.  $15.40. 

11.  An  importer  received  a  quantity  of  Leghorn  hats,  invoiced 
at  25256  lire  16  soldi  which  was  paid  in  U.  S.  gold  coin,  ex- 
ported at  a  cost  of  3  %  for  transportation  and  insurance,  the  price 
of  fine  gold  in  Leghorn  being  131  lire  per  ounce  Troy.  How 
much  less  would  the  goods  have  cost  in  store,  had  payment  been 
made  by  draft  on  Leghorn,  purchased  at  the  rate  of  16  cents  per 
lira?  Am.  §64.0 1. 

NoTK.  — In  U.S.  gold  coiqage,  $10  contains  258  X  .9  =  232.2  grains  of//ie 
gold,  (601). 

12.  When  silver  is  worth  in  England  67d.  per  oz.  fine,  what 
sum  of  money  in  the  U.  S.  silver  coinage  of  1853  is  equal  to  20 
shillings,  or  £1  sterling?  Am.  $1,975. 

13.  At  what  rate  of  premium  is  Prussian  coin,  when  S88.23  in 
U  S.  silver  coinage  of  1837  is  paid  for  125  thalers  ?    Am.  2  %. 


348  PERCENTAGE. 

ARBITRATED   EXCHANGE. 

607,  Arbitration  of  Exchange  is  the  process  of  computing 
exchange  between  two  places  by  means  of  one  or  more  interme- 
diate exchanges. 

Notes. —  1.  When  there  is  only  one  interraediate  exchange,  the  process  is 
called  Simple  Arbitration  ;  when  there  are  two  or  more  interaiediato  exchanges, 
the  process  is  called  Comjoound  Arbitration. 

2.  The  arbitrated  price  is  generally  either  greater  or  less  than  the  price  of 
direct  exchanges;  and  the  object  of  arbitration  is  to  ascertain  the  best  route  for 
making  drafts  or  remittances. 

608.  There  are  always  three  methods  of  receiving  money 
from  a  place,  or  of  transmitting  money  to  a  place,  by  means  of 
indirect  exchange  through  one  intervening  place.     Thus, 

If  A  is  to  receive  money  from  C  through  B, 
1st.  A  may  draw  on  B,  and  B  draw  on  C ; 
2d.  A  may  draw  on  B,  and  C  remit  to  B ; 
3d.  B  may  draw  on  C,  and  remit  to  A. 

If  A  is  to  transmit  money  to  C  through  B, 
1st.  A  may  remit  to  B,  and  B  remit  to  C ; 
2d.  A  may  remit  to  B,  and  C  draw  on  B ; 
8d.  B  may  draw  on  A,  and  remit  to  C. 

1.  A  man  in  Albany,  N.  Y.,  paid  a  demand  in  Paris  of  5400 
francs,  by  remitting  to  Amsterdam  at  the  rate  of  21  cents  for  10 
stivers,  and  thence  to  Paris  at  the  rate  of  28  stivers  for  3  francs ; 
how  much  Federal  money  was  required  ? 

OPERATION.  Analysis.     We  are  to  deter- 

$  f  ?  ^  =  5400  francs.  mine  how  much  Federal  money 

3  francs    =  28  stivers.  is  equal  to  5400  francs,  and  the 

10  stivers  =  821.  question    may    bo    represented 

(  ?)  =  ?51058.40,  Arts.    "^"«  =  ^  (  0  -  5400  francs.  Now 

Or, 


10 


5400 


since  3  francs  =  28  stivers,  and 

10  stivers  =  $.21,  we  know  that 

if  the  required   sum  be  multi- 

^^  plied    successively  by  3   francs 

LLl! and  10  stivers,  the  result  will  be 

'.    ()  =  §1058.40,  Ans.        equal  to  the  product  of   5400 

francs  by  28  stivers  and  $.21 
successively,  (Ax.  3).  Canceling  the  units  of  currency,  1  franc,  1 
stiver,  and  $1,  and  also  the  equal  numerical  factors,  we  have  (?) 
=  $1058.40,  the  sum  required. 


EXCHANGE.  849 

Or,  since  the  course  of  exchange  betTveen  Amsterdam  and  Paris 
gives  1  franc  =  ^j  stivers,  and  the  course  between  Albany  and  Am- 
sterdam gives  1  stiver  =  f  J  cents,  vre  multiply  the  5400  francs  by  ^^ 
and  f  J-  successively,  using  the  vertical  line  and  cancellation,  and  obtain 
$1058.40,  as  before. 

Note. —  In  the  first  statement  the  rates  of  exchange  are  so  arranged  that  the 
same  unit  of  currency  shall  stand  on  opposite  sides  in  each  two  consecutive 
equations,  in  order  that  these  factors  may  all  be  canceled. 

2.  A  resident  of  Naples  having  a  bequest  of  $8720  made  him 
in  Boston,  orders  the  remittance  to  be  made  to  his  agent  in  Lon- 
don, who  remits  the  proceeds  to  Naples,  reserving  his  commission 
of  ^  %  on  the  draft  sent.  If  exchange  on  London  is  9  %  in 
Boston,  and  the  rate  between  London  and  Naples  is  £1  for  5 
scudi,  how  much  does  the  man  realize  from  his  bequest  ? 

oPERATioi  Analysis.      AVe   make 

(?)  scudi  =  $8720  the  statement  as  in   the 

$40  X  1.09  =  £9  first   example,   according 

£1  =  5  scudi.  to  the  given  rates  of  ex- 

1.005  change.     Then,  since  the 

(?)  =  8955  scudi  3  carlini.      ^g^^*  '\  *^  ^^^^«*  i  /^ 

commission   on  the    face 

of  the  draft  before  the  purchase,  vre  place  1.005  on  the  left  as  a 
divisor,  (159),  and  obtain  by  cancellation  8955  scudi  3  carlini  as  the 
proceeds  of  the  exchange. 

Note. — Since  the  par  of  exchange  on  Ennrland  is  £9  =  $40,  the  course  of  ex- 
change will  always  be  £9  =  $40  X  1  plus  the  rate  of  exchange. 

3.  A  merchant  in  Chicago  directs  his  agent  in  Albany  to  draw 
upon  Baltimore  at  1  %  discount,  for  $1200  due  from  the  sales 
of  produce ;  he  then  draws  upon  the  Albany  agent  at  2  %  pre- 
mium, for  the  proceeds,  after  allowing  the  agent  to  reserve  J  % 
for  his  commission.  What  sum  does  the  merchant  realize  from 
his  produce  ? 

OPERATION".  Analysis.     According  to  the 

(?)  C.  =   1200  B.  given  rates  of  exchange,  100  dol- 

100  B.  =        99  A.  lars  in  Baltimore  is  equal  to  99 

100  A.  =      102  a  dollars  in  Albany;  and  100  dol- 

.995  lars  in  Albany  is  equal  to  102 

/'?\__  §1205.70    Ans.        dollars  in  Chicago  ;  and  since  the 

unit  of  currency  is  the  same  in 

<3ach  place,  being  $1,  we  represent  its  exchange  value  in  each  town 


350  PERCENTAGE.* 

"by  the  initial  letter,  and  make  the  statement  as  in  the  other  exam- 
ples. Then,  since  the  agent  is  to  reserve  J  %  commission  from  the 
avails  of  his  draft,  we  place  1  —  .005  =  .995  on  the  right  as  a  mul- 
tiplier, and  obtain  by  cancellation  { ?  )  =  $1205.70,  the  answer. 
From  these  principles  and  illustrations  we  have  the  following 
Rule.  I.  Represent  the  required  sum  hy  ( 'i  ),  with  the  proper 
unit  of  currency  affixed,  and  place  it  equal  to  the  given  sum  on  the 
right. 

II.  Arrange  the  given  rates  of  exchange  so  that  in  any  two  con- 
secutive equations  the  same  unit  of  currency  shall  stand  on  op^posite 
sides. 

III.  When  there  is  commission  for  drawing,  place  1  minus  the 
rate  on  the  left  if  the  cost  of  exchange  is  required,  and  on  the  right 
if  proceeds  are  required ;  and  when  there  is  commission  for  remit- 
ting, place  1  plus  the  rate  on  the  right  if  cost  is  required,  and  on 
the  left  if  proceeds  are  required. 

lY.  Divide  the  product  of  the  numbers  on  the  right  hy  the  j^^'od- 
uct  of  the  numhers  on  the  left,  cancelling  equal  factors ',  the  result 
will  he  the  answer. 

Notes.  —  1.  Commission  for  drawing  is  commission  on  the  sale  of  a  draft; 
commission  for  remitting  is  commission  on  the  purchase  price  of  a  draft. 

2.  The  above  method  is  sometimes  called  the  Chain  liulef  or  Conjoined  Pro- 
portion. 

EXAMPLES   FOR   PRACTICE. 

1.  A  gentleman  in  Philadelphia  wishes  to  deposit  $5000  in  a 
bank  at  Stockholm,  by  remitting  to  Liverpool  and  thence  to  Stock- 
holm ;  if  exchange  on  Liverpool  is  at  10  %  premium  in  Phila- 
delphia, and  the  course  between  Liverpool  and  Stockholm  is  6 
roubles  48  copecks  per  £1,  how  much  money  will  the  man  have 
in  bank  at  Stockholm,  allowing  the  agent  at  Liverpool  J  %  for 
remitting?  A7is^  6610  roubles  74  copecks. 

2.  When  exchange  at  New  York  on  Paris  is  5  francs  16  cen- 
times per  $1,  and  at  Paris  on  Hamburg  2i  francs  per  marc  banco, 
what  will  be  the  arbitrated  price  in  New  York  of  7680  marc 
bancos  of  Hamburg?  Ans.  $3162.79. 

3.  A  man  in  Cleveland  wishes  to  draw  on  New  Orleans  for  a 
bank  stock  dividend  of  $750,  and  exchange  direct  on  New  Or- 
leans is  li    %  discount;  how  much  will  he  save  by  drawing  on 


.      .  EXCHANGE.  35I 

his  agent  in  New  York  at  1^  %  premium,  allowing  his  agent  ta 
draw  on  New  Orleans  at  1  %  discount,  brokerage  at  ^  %  '/ 

4.  A  gentleman  in  Boston  drew  on  Wurtemberg  for  GO 00  gild- 
ers at  $.415  per  gilder;  how  much  more  would  he  have  received 
if  he  had  ordered  remittance  to  London,  and  thence  to  New  York, 
exchange  at  AYurtemberg  on  London  being  11 J  gilders  per  £1, 
and  at  London  on  New  York  9 J  %,  in  favor  of  sterling,  broker- 
age at  IJ  %  in  London  for  remitting?  Ans.  $67.66. 

5.  If  at  Philadelphia  exchange  on  Liverpool  is  at  9|  %  pre- 
mium, and  at  Liverpool  on  Paris  26  francs  8'6  centimes  per  £1 ; 
what  is  the  arbitrated  course  of  exchange  between  Philadelphia 
and  Paris,  through  Liverpool  ?  Ans,  1  franc  =  $.181. 

6.  An  American  resident  of  Amsterdam  wishing  to  obtain 
funds  from  the  U.  S.  to  the  amount  of  $6400,  directs  his  agent 
in  London  to  draw  on  the  U.  S.  and  remit  the  proceeds  to  him  in 
a  draft  on  Amsterdam,  exchange  on  the  U.  S.  be'ing  at  8  %  in 
favor  of  London,  and  the  course  between  London  and  Amsterdam 
being  18d.  per  gilder.  If  the  agent  charges  commission  at  J  % 
both  for  drawing  and  remitting,  how  much  better  is  this  arbitra- 
tion than  to  draw  directly  on  the  U.  S.  at  40  cents  per  gilder  ? 

7.  A  speculator  in  Pittsburgh,  having  purchased*  58  shares  of 
railroad  stock  in  New  Orleans,  at  95  %,  remits  to  his  agent  in 
New  York  a  draft  purchased  at  2  %  premium,  with  orders  for  the 
agent  to  remit  the  sum  due  in  N.  0.  Now,  if  exchange  on  N.  0. 
is  at  i  (fo  discount  in  N.  Y.,  and  the  agent's  commission  for  re- 
mitting is  i  %,  how  much  does  the  stock  cost  in  Pittsburgh? 

Ans.  $5606.08. 

8.  A  banker  in  New  York  remits  $3000  to  Liverpool,  by  arbi- 
tration, as  follows  :  first  to  Paris  at  5  francs  40  centimes  per  $1 ; 
thence  to  Hamburg  at  185  francs  per  100  marcs;  thence  to  Am- 
sterdam at  85  stivers  per  2  marcs;  thence  to  Liverpool  at  220 
stivers  per  £1  sterling.  How  much  sterling  money  will  he  have 
in  bank  at  Liverpool,  and  what  will  be  his  gain  over  direct  ex- 
change at  10  %  premium  ? 

.         (  Proceeds  in  Liverpool,  £696  lis.  2d. 
I  Gain  by  arbitration^         £82  18s.  5d. 


852  PERCENTAaB. 

EQUATION  OF  PAYMENTS. 

609.  Equation  of  Payments  is  the  process  of  finding  the 
mean  or  equitable  time  of  payment  of  several  sums,  due  at  dif- 
ferent times  without  interest. 

610.  The  Term  of  Credit  is  the  time  to  elapse  before  a  debt 
becomes  due. 

611.  The  Average  Term  of  Credit  is  the  time  to  elapse  before 
several  debts,  due  at  different  times,  may  all  be  paid  at  once,  with- 
out loss  to  debtor  or  creditor. 

615.  The  Equated  Time  is  the  date  at  which  the  several 
debts  may  be  canceled  by  one  payment. 

613.  To  Average  an  Account  is  to  find  the  mean  or  equit- 
able time  of  payment  of  the  balance. 

614:.  A  Focal  Date  is  a  date  with  which  all  the  others  are  com- 
pared in  averaging  an  account. 

Note.  —  Each  item  of  a  book  account  draws  interest  from  the  time  it  is  due, 
which  may  be  either  at  the  date  of  the  transaction,  or  after  a  specified  term  of 
credit. 

In  averaging,  there  are  two  kinds  of  equations,  Simple  and 
Compound.    ' 

61^.  A  Simple  Equation  is  the  process  of  finding  the  aver- 
age time  when  the  payments  or  account  contains  only  one  side, 
which  may  be  either  a  debit  or  credit. 

616.  A  Compound  Equation  is  the  process  of  averaging 
when  both  debts  and  credits  are  to  be  considered. 

SIMPLE    EQUATIONS. 
CASE    I. 

617.  When  all  the  terms  of  credit  begin  at  the  same 
(late. 

1.  In  settling  with  a  creditor  On  the  first  day  of  April,  I  find 
that  I  owe  him  $12  due  in  5  months,  815  due  in  2  months,  and 
$18  due  in  10  months ;  at  what  time  may  T  pay  the  whole  amount? 


EQUATION  OF  PAYMENTS.  353 

OPERATION.  Analysis.     The 

$12  X     5  =    GO  whole   amount   to    be 

I  15  X     2  =     30  paid,  as  seen  in  the  ope- 

18x10  =  180  ration,  is  $45  ;  and  we 

L^K       ©T^  270  ^^®  *^  ^^^  ^^^  ^^^S  ^* 

B        270  --45  =  6  mo.,  average  credit,         shall  be  withheld,  or 
IB        Apr.  1,  +  6  mo.  =  Oct.  1,  Ans.  what  term  of  credit  it 

I^P  shall  have,  as  an  equiv- 

alent for  the  various  terms  of  credit  on  the  different  items.  Now 
the  value  of  credit  on  any  sum  is  measured  by  the  product  of  the 
money  and  time.  Therefore,  the  credit  on  $12  for  5  mo.  =  the  credit 
on  $60  for  1  mo.,  because  12  X  5  =  GO  X  1.  In  like  manner,  we  have 
the  credit  on  $15  for  2  mo.  =  the  credit  on  $30  for  1  mo. ;  and  the 
credit  on  $18  for  10  mo.  =  the  credit  on  $180  for  1  mo.  Hence,  by 
addition,  the  value  of  the  several  terms  of  credit  on  their  respective 
sums  equals  a  credit  of  1  month  on  $270 ;  and  this  equals  a  credit  of 
G  months  on  $45,  because  45  X  6  =  270  x  1.     Hence  the  following 

Rule.  I.  Multiply  each  payment  hy  its  term  of  creditj  and 
divide  the  sum  of  the  products  hy  the  sum  of  the  payments ;  the 
quotient  loill  he  the  average  term  of  credit, 

II.  Add  the  average  term  of  credit  to  the  date  at  which  all 
the  credits  begin  ;  the  result  will  he  the  equated  time  of  payment. 

Notes.  —  1.  The  periods  of  time  used  as  raultipliers  must  all  be  of  the  same 
deuomination,  and  the  quotient  will  be  of  the  same  denomination  as  the  terms 
of  credit;  if  these  be  months,  and  there  be  a  remainder  after  the  division,  con- 
tinue the  division  to  days  by  reduction,  always  taking  the  nearest  unit  in  the  last 
result. 

2.  The  several  rules  in  equation  of  payments  are  based  upon  the  principle  of 
bank  discount;  for  they  imply  that  the  discount  of  a  sum  paid  before  it  is  duo 
equals  the  interest  of  the  same  amount  paid  after  it  is  due. 

EXAMPLES   FOR   PRACTICE. 

1.  On  the  first  day  of  January,  1860,  a  man  gave  3  notes,  the 
first  for  §500  payable  in  30  days ;  the  second  for  $400  payable  in 
60  days;  the  third  for  $600  payable  in  90  days.  What  was  the 
average  term  of  credit,  and  what  the  equated  time  of  payment  ? 

Ans    Term  of  credit,  62  da. ;  time  of  payment,  Mar.  3,  1860. 

2.  A  man  purchased  real  estate,  and  agreed  to  pay  i  of  the  price 
in  3  mo.,  }  in  8  mo.,  and  the  remainder  in  1  year.  Wishing  to 
cancel  the  whole  obligation  at  a  single  payment,  how  long  shall 
this  payment  be  deferred  ?  ^ 

30*  X 


854 


PERCENTAGE. 


3.  I  owe  $480  payable  in  90  days,  and  $320  payable  in  60  days. 
My  creditor  consents  to  an  extension  of  time  to  1  year,  and  oficrs 
to  take  my  note  for  the  whole  amount  on  interest  at  6  per  cent, 
from  the  equated  time,  or  a  note  for  the  true  present  worth  of 
both  debts,  on  interest  from  date.  How  much  will  I  gain  if  I 
choose  the  latter  condition?  Ans,  $1.14. 

4.  Bought  merchandise  April  1,  as  follows:  $280  on  3  mo., 
$300  on  4  mo.,  $200  on  5  mo.,  $560  on  6  mo. ;  what  is  the 
equated  time  of  payment?  Ans.  Aug.  24. 


CASE   II. 

618*    "WTien  the  terms  of  credit  begin  at  different 
dates. 

1.  When  does  the  amount  of  the  following  bill  become  due, 
per  average  ? 

Charles  Crosby, 

1860.  To  Bronson  &  Co.,  Dr. 

Jan.  12.     To  Mdse., $400 

"    16.      ^'   Mdse.  on  2  mo., 600 

Apr.20.      "   Cash, 375 


FIRST    OPERATION. 


SECOND    OPERATION. 


Due      Da, 

Items. 

Prod. 

Jan.  12  i 
Mar.  16  1  64 
Apr.  20   99 

400 
600 
375 

38400 
37125 

1375 

75525 

Due. 

Da. 
99 

35 
0 

Items. 

Prod. 

Jan.  12 
Mar.  16 
Apr.  20 

400 
600 
375 

1375 

39600 
21000 

60600 

Ans 


75525  -^  1375  ==  55  da. 
55  da.  after  Jan.  12, 
or  Mar.  7. 


•!: 


Ansig  < 


60600  -f  1375  =  44  da. 
44  da,  before  Apr.  20, 
or  Mar.  7. 


Analysis.  The  three  items  of  the  bill  are  due  Jan.  12,  Mar.  16, 
and  Apr.  20,  respectively.  In  the  first  operation  we  use  the  earliest 
maturity,  Jan.  12,  for  a  focal  date,  and  find  the  difi'erence  in  days 
between  this  date  and  each  of  the  others ;  thus,  from  Jan.  12  to  Mar, 


I 


EQUATION  OF  PAYMENTS.  355 


16  is  64  da. ;  from  Jan.  12  to  Apr.  20  is  99  da.  Hence,  from  Jan.  12 
the  first  item  has  no  credit,  the  second  lias  64  days'  credit,  and  the 
third  99  days'  credit,  as  appears  in  the  column  marked  da.  We  now 
proceed  to  find  the  products  as  in  Case  I,  whence  we  obtain  the  ave- 
rage credit,  55  da.,  and  the  equated  time,  Mar.  7. 

In  the  second  operation,  the  latest  maturity.  Apr.  20,  is  taken  for  a 
focal  date,  and  the  work  may  be  explained  thus :  Suppose  the  account 
to  be  settled  Apr.  20.  At  that  time  the  first  item  has  been  due  99 
days,  and  must  therefore  dj^-aw  interest  for  this  time.  But  interest 
on'$400  for  99  days  =  the  interest  on  $39600  for  1  day.  The  second 
item  must  draw  interest  35  days ;  but  interest  on  $600  foi  35  days  ^= 
interest  on  $21000  for  1  day.  Taking  the  sum  of  the  products,  we  find 
that  the  whole  amount  of  interest  due  Apr.  20  equals  the  interest  on 
$60600  for  1  day ;  and  this  is  found,  by  division,  equal  to  the  interest 
on  $1375  for  44  da.,  which  is  the  average  term  of  interest.  Hence 
the  account  would  be  settled  Apr.  20,  by  paying  §1375,  with  interest 
on  the  same  for  44  days.  This  shows  that  the  $1375  has  been  used 
44  days,  that  is,  it  falls  due  Mar  7,  without  interest.  Hence  we  have 
the  following 

Rule.  I.  Find  the  time  at  which  each  item  Lecomes  due,  hy 
adding  to  the  date  of  each  transaction  the  term  of  credit y  if  any  he 
specifiedj  and  write  these  dates  in  a  column. 

II.  Assume  either  the  earliest  or  the  latest  date  for  a  focal  date^ 
and  find  the  difference  in  days  between  the  focal  date  and  each  of 
the  other  dates,  and  write  the  results  in  a  second  column. 

III.  Write  the  items  of  the  account  in  a  third  column,  and  mul- 
tiply each  hy  the  corresponding  number  of  days  in  the  preceding 
columii,  writing  the  products  in  a  fourth  column. 

IV.  Divide  the  sum  of  the  products  hy  the  sum  of  the  items. 
TJie  quotient  will  he  the  average  term  of  credit  or  interest,  and 
must  be  reckoned  from  the  focal  date  TOWARD  the  other  dates,  to 
find  the  equated  time  of  payment. 

Notes.  —  1.  When  dollars  and  cents  are  given,  it  is  generally  suflficient  to  take 
only  dollars  in  the  multiplicand,  rejecting  the  cents  when  less  than  50,  and  car- 
ryinj^  1  to  the  dollars,  if  the  cents  are  more  than  50. 

2.  Months  in  any  terms  of  credit  are  understood  to  be  calendar  months;  the 
time  must  therefore  be  carried  forward  to  the  same  day  of  the  month  in  which 
the  term  of  credit  expires. 


356  percentage. 

examples  ror  practice. 

1.  James  Gordon, 

1860.                                        To  Henry  Lancey,  Dr. 
Mar.    4.     To    100  yd.  Cassimere,      @  $2  50, S250 

"      25.      "  300C    "    French  Prints,"       .12, 360 

Apr.  16.      "  1200    "    Sheeting,         "       .08, 96 

"     30.      "     400    "    Oilcloth,        «      .50, 200 

May  17.      "     Sundries, 350 

When  is  the  above  bill  due,  per  average  ? 

Arts.  Apr.  12,  1860. 

2.  I  sell  goods  to  A  at  different  times,  and  for  different  terms 
of  credit,  as  follows : 

Sept.  12,  1859,  a  bill  on  30  days^  credit,  for  $180 

Oct.      7,     "         "  30  "  "  300 

Nov.  16,      "         "  60  "  "  150 

Dec.   20,      ''  "  90  "  "  350 

Jan.    25,  1860,  "  30  "  "  130 

Feb.   24,     "  ''  30  "  "  140 
Tf  I  take  his  note  in  settlement,  at  what  time  shall  interest 
commence  ? 

3.  What  is  the  average  of  the  following  account? 

1860,  Oct.    1.     Mdse.,  on  60  da.,.. ..,....„ $240 

"       Nov.12.        "         "       ''       500 

"       Dec.  25.        "         "       " 436 

1861,  Jan.  16.        "         "       "       325 

"       Feb.  24.        "         "       "       436 

''       Mar.I7.        "         "       "       537 

Ans.   Mar.  10,  1861. 

4.  I  have  4  notes,  as  follows :  the  first  for  $350,  due  Aug.  16, 
1859  •  the  second  for  $250,  due  Oct.  15, 1859 ;  the  third  for  $300, 
due  Dec.  14,  1859;  the  fourth  for  $248,  due  Feb.  12,  1860. 
When  shall  a  note  for  which  I  may  exchange  the  four,  be  made 
payable  ? 


I 


m 


EQUATION  OF  PAYxMENTS. 
COMPOUND    EQUATIONS. 

C19.    1.  Average  the  following  account. 
John  Lyman. 


357 


Or. 


1860. 

1860. 

June  12 

To  Mdse. 

530 

GO 

June  24 

By  draft  at  30  da. 

480 

00 

Sept.  12 

li            u 

428 

00 

Aug.   20 

"    cat?h, 

280 

00 

pet.     28 

"   Sundries, 

440 

00 

Oct.       8 

a       ii 

140 

00 

OPERATION. 


Dr. 


Due. 

Da. 

Items. 

Products. 

Due. 

Da. 

Items. 

Products. 

June  12 

Sept.  12 
Oct.     28 

138 

46 

0 

530 
428 
440 

73140 
19688 

July  27 
Aug.  20 
Oct.     8 

93 

09 
20 

480 
230 
140 

44640 

15870 

2800 

1398 
850 

92828 
63310 

850 

63310 

' 

Balances, 

548 

29518 

29518  -T-  548  =  54  da.,  average  term  of  interest. 
Oct.  28  —  54  da.  =  Sept.  4,  balance  due. 

Analysis. — In  this  operation  we  have  written  the  dates  of  maturity 
on  either  side,  allowing  3  days'  grace  to  the  draft.  The  latest  date, 
Oct.  28,  is  assumed  as  the  focal  date  for  botJi  sides,  and  the  two  columns 
marked  da.  show  the  difference  in  days  between  the  focal  date  and 
each  of  the  other  dates.  The  products  are  obtained  as  in  simple 
equations,  and  the  balance  found  between  the  items  on  the  two  sides, 
and  also  between  the  products.  These  balances,  being  both  on  the 
Dr.  side,  show  that  there  is  due  on  the  day  of  the  focal  date,  $548, 
with  interest  on  $29518  for  1  day.  By  division,  this  interest  is  found 
to  be  equal  to  the  interest  on  $548  for  54  days.  Hence  this  balance, 
$548,  has  been  due  54  days ;  and  reckoning  back  from  the  focal  date, 
we  obtain  the  equated  time  of  payment,  Sept.  4. 

Had  we  taken  the  earliest  maturity,  June  12,  for  the  focal  date,  we 
should  have  obtained  84  days  for  the  interval  of  time ;  and  since  in 
this  case  the  products  would  represent  the  credit  to  which  the  several 
items  are  entitled  after  June  12,  we  should  add  84  days  to  the  focal 
date,  which  would  give  Sept.  4,  as  before. 

2.  When  is  the  balance  of  the  following  account  due,  per 
average  ? 


358 


PERCENTAGE. 
Cliarles  Derby. 


1859. 

1859. 



Jan.     21 

To  Mdse. 

H2 

00 

Jan.       1 

By  cash, 

84 

00 

Mar.      5 

"           u 

145 

00 

Feb.       4 

40 

00 

"    22 

"       " 

194 

00 

Mar.     .30 

"      " 

12 

00 

OPERATION. 


Cr. 


Due. 

Da. 

Items. 

Products. 

Due. 

Da. 

Items. 

Products. 

Jan.    21 

Mar.      5 

«      22 

68 
25 
8 

32 
145 
194 

2176 
3625 
1552 

Jan.      1 
Feb.      4 
Mar.   30 

88 
54 
0 

84 
40 
12 

7392 
2160 

371. 
136 

7353 

136 

9552 
7353 

Balance  of  account, 

235 

Balance  of  products, 

2199 

2199  -T-  235  =  9  da. ;  Mar.  30+9  da.  =  Apr.  8,  Ans. 
Analysis.  We  take  the  latest  maturity,  Mar.  30,  for  the  focal  date, 
and  consequently  the  products  represent  the  interest  due  upon  the 
several  items,  at  that  date.  We  find  the  balance  of  the  items  upon 
the  Dr.  side,  and  the  balance  of  the  products  upon  the  Cr.  side.  The 
-debtor  therefore  owes,  on  Mar.  30,  $235,  but  is  entitled  to  such  a  term 
of  interest  on  the  same  as  will  be  equivalent  to  the  interest  on  $2199 
for  1  day,  which  by  division,  is  found  to  be  9  da.  Hence  the  balance 
is  due  Mar.  30+9  da.  =  Apr.  8.  Thus  we  see  that  when  the  balances 
are  on  opposite  sides,  the  interval  of  time  is  counted  from  the  other 
dates.  If  we  take,  in  this  example,  the  earliest  date  for  the  focal  date, 
the  balances  will  both  be  upon  the  Dr.  side,  and  the  interval  of  time 
will  be  97  da.,  which  reckoned  forward  from  the  focal  date,  will  give 
the  equated  time  as  before. 

6S0*    From  these  examples  we  derive  the  following 
lluLE.     I.   Find  the  time  when  each  item  of  the  account  is  due, 
and  write  the  dates,  in  two  columns,  on  the  sides  of  the  account  to 
which  they  respectively  belong. 

II.  Use  either  the  earliest  or  the  latest  of  these  dates  as  the  focal 
date  for  both  sides,  ana  find  the  products  as  in  the  last  case. 

III.  Divide  the  balance  of  the  products  by  the  balance  of  the 
account ;  the  quotient  will  be  the  interval  of  time,  which  must  be 
reckoned  from  the  focal  doie  TOWARD  the  other  dates  when  both 


EQUATION  OF  PAYMENTS. 


359 


^^ktes  when  the  balances  are  on  opposite  sides  of  the  account, 

^^^J^'oTES.  —  1.  Instead  of  the  products,  we  may  obtain  the  interest,  at  any  per 
cent.,  on  the  several  items  for  the  corresponding  intervals  of  time,  and  divide 
the  balance  of  interest  by  the  interest  on  the  balance  of  the  account  for  1  day  ; 
the  quotient  will  be  the  interval  of  time  to  be  added  to,  or  subtracted  from  the 
focal  date,  according  to  the  rule.  The  time  obtained  will  be  the  same,  at  what- 
ever rate  the  interest  be  computed. 

2.  There  may  be  such  a  combination  of  debits  and  credits,  that  the  equated 
time  will  be  earlier  or  later  than  any  date  of  the  account.  ^ 


EXAMPLES   FOR   PRACTICE. 

1.  Required,  the  average  maturity  of  the  following  account. 

A.  Z.  Armour, 


Dr. 

Cr. 

1859. 

1859. 

Feb.      12 

To  Mdse. 

85 

75 

March  15 

By  bal.  old  acc't. 

97 

36 

25 

((          u 

36 

24 

April    17 

"   cash, 

56 

(K) 

April    16 

((       « 

174 

96 

May      25 

a       u 

25 

00 

Mav      20 

«       « 

94 

78 

June       8 

"   sundri'es, 

94 

75 

OPERATION. 


Dr. 


Cr. 


Due. 

Da. 

Items. 

Int. 

Due. 

Da. 

Items. 

Int. 

1.3S 
.40 
.06 

Feb.    12 

"        25 
April  16 
Ma^    20 

116 
103 
53 
19 

85.75 

36.24 

17496 

94.78 

1.66 

.62      1 
1.55 
.30 

March  15 
April    17 
May      25 
June      8 

85 
62 
14 

97.36 
56.00 
25.00 
94.75 

391.73 
273.11 

4.13 
1.93 

273.11 

1.93 

Balances, 

118.62 

2.20 

Int  on  $118.62  for  1  da.  =  $.0198. 
2.20---.0198=lll  da.;  June  8—111  da.=:.Feb.  17,1859,  Ans. 

Analysis.  Taking  the  latest  maturity,  June  8,  for  the  focal  date, 
we  find  the  interest  of  each  item,  at  6  ^,  from  its  maturity  to  the 
focal  date ;  then,  taking  the  balance,  we  find  the  interest  due  on  the 
account  to  be  $2.20.  Dividing  this  interest  by  the  interest  on  the 
balance  of  the  items  for  1  day,  we  obtain  111  da.,  the  time  required 
for  the  interest,  $2.20,  to  accrue.  The  average  maturity,  therefore, 
is  June  8  —  111  da.  =  Feb.  l7,  1859. 

It  is  evident  that  when  the  balances  occur  on  opposite  sides,  the 
interval  of  time  will  be  reckoned  as  in  the  method  by  products. 


860 


PEllCENTAQE. 


2.  What  is  the  balance  of  the  following  account, ^.and  when  is 

it  due  ?  ■■         ^ 

Thomas  Lardner, 

Dr.  Cr. 


1860. 

1860. 



March  1 

To  Sundries, 

436 

00 

March  25 

By  draft,  at  60  da. 

400 

00 

April  12 

"  Mdse. 

548 

00 

April      6 

"      «          30  '• 

650 

00 

July    16 

it         u 

312 

00 

June    20 

"   cash. 

200 

00 

Sept.  14 

"      " 

536 

00 

Aug.      3 

"      " 

84 

00 

Ans.  Balance,  $498;  due  June  22,  1860. 
3.  When  shall  a  draft  for  the  settlement  of  the  following  ac- 
count  be  made  payable  ? 

David  Sanford. 


Dr. 

Cr. 

1859. 

1869. 

Jan.        1 

To  Mdse.  on  3  mo. 

54 

i 

April    1 

By  cash, 

50 

•00 

Feb.      12 

"       "       '^  2     " 

28 

May    16 

"  draft,  at  30  da. 

30 

00 

March  16 

"   Sundries, 

95 

75 

June  12 

ii        a 

125 

00 

June     25 

"   Mdse. 

26 

32 

"      20 

"   cash, 

150 

00 

Ans.  Aug.  28,  1859. 


Oliver  Waimcright. 


Dr. 

Cr. 

1858. 

1858. 

Jan.        1 

To  Mdse. 

36 

72 

Jan.      10 

By  fcash, 

98 

72 

Feb.        1 

U           ii 

48 

25 

"         21 

it       a    ' 

25 

84 

March  17 

((       ii 

72 

36 

March  23 

"   sundries, 

15 

17 

April      1 

ii      il 

98 

48 

April      6 

a          a           ' 

8 

90 

If  the  above  account  were  settled  April  6,  1858,  by  draft  on 
time,  how  many  days'  credit  should  be  given  ?  Ans.  20  Qa. 

6.  I  owe  $1000  due  Apr.  25.    If  I  pay  $560  Apr.  1,  and  §324 
Apr.  21,  when,  in  equity,  should  I  pay  the  balance  ? 

Ans.  Aug.  30. 

Note. — Make  the  $1000  the  Dr.  side  of  an  account,  and  the  payments  the  Cr. 
Bide,  and  then  average. 


6.  A  man  owes  $684,  payable  Aug.  12,  and  $4:68,  payable  Oct. 
15.  If  he  pay  $839;  Aug.  1,  what  will  be  the  equated  time  for 
the  payment  of  the  balance  ?  Ans.  Dec.  15. 

7o  A  man  holds  3  notes,  the  first  for  $500,  due  March  1,  the 
second  for  $800,  due  June  1,  and  the  third  for  $600,  due^^ug.  1. 
He  wishes  to  exchange  them  for  two  others,  one  of  which  shall 
be  for  $1000,  payable  Apr.  1 ;  what  shall  -be  the  face  and  when 
the  maturity  of  the  other  ? 

Ans.  Face,  $900 ;  maturity,  July  28. 


EQUATION  OF  PAYMENTS. 


861 


8.  A  owes  $500,  due  Apr.  12,  and  $1000,  due  Sept.  20,  and 
wishes  to  discharge  the  obligation  by  two  equal  payments,  made 
at  an  interval  of  60  days;  when  must  the  two  payments  be  made  ? 

Ans.   1st,  June  28;  2d,  Aug.  27. 

9.  When  shall  a  note  be  made  payable,  to  balance  the  followino- 

account  ? 

James  Tyler, 


Dr. 

Cr. 

1«59. 

1859. 

June   12 

To  Mdflo.  on  3  mo. 

530 

84 

Sept.  14 

By  caFh, 

436 

no 

*'       20 

"        "     f'     " 

236 

48 

'•    25 

i.           u 

320 

00 

«       30 

a          a     «     « 

V39 

56 

Oct.     3 

((      il 

560 

00 

July      6 

«          a      u      a 

273 

44 

"  n 

li          u 

370 

00 

"       16 

"          <;      a      « 

194 

78 

Nov.  16 

a        « 

840 

00 

"       29 

u          u      a     it 

636 

42 

*•    24 

<i        i( 

5(0 

00 

10.  I  received  goods  from  a  wholesale  firm  in  New  York,  in 
parcels,  as  per  bills  received,  namely :  Apr.  1,  a  bill  for  $536.78; 
May  16,  $2156.94;  June  12,  $843.75;  July  12,  $594.37;  Sept. 
18,  $856.48.  In  part  payment,  I  remitted  cash  as  follows:  June 
8,  $500;  July  1,  $1000;  Nov.  1,  $1500.  When  is  the  balance 
payable,  allowing  credit  of  2  months  for  the  merchandise  ? 

Ans.  July  23 > 

ACCOUNT   SALES. 

©SI.  An  Account  Sales  is  an  account  rendered  by  a  commis- 
sion merchant  of  goods  sold  on  account  of  a  consignor,  and  con- 
tains a  statement  of  the  sales,  the  attendant  charges,  and  the  net 
proceeds  due  the  owner. 

03S.  Guaranty  is  a  charge  made  in  addition  to  commission, 
for  securing  the  owner  against  the  risk  of  non-payment,  in  case  of 
goods  sold  on  credit. 

6S3.  Storage  is  a  charge  made  for  keeping  the  goods,  and 
may  be  reckoned  by  the  week  or  month,  on  each  article  or  piece. 

G24.  Primage  is  an  allowance  paid  by  a  shipper  or  consignor 
of  goods  to  the  master  and  sailors  of  a  vessel,  for  loading  it. 

OS«i.    A    commission   merchant   having  sold   a  shipment  of 
goods  by  parts  at  different  times,  and  on  various  terms,  makes  a 
final  settlement  by  deducting  all  charges,  and  accrediting  the  owner 
with  the  net  proceeds.     It  is  evident,  therefore, 
31 


362 


PERCENTAGE. 


I.  That  commission  and  guaranty  should  be  accredited  to  the 
agent  at  the  average  maturity  of  the  sales, 

II.  That  the  net  proceeds  should  be  accredited  to  the  con- 
signor at  the  average  maturity  of  the  entire  account. 

Hence  the  following 

Rule.  I.  To  compute  the  storage.  —  Multiple/  each  article  or 
parcel  hy  the  time  it  is  in  store j  and  multiply  the  sum,  of  the  pro- 
ducts hy  the  rate  per  unit ;   the  result  will  he  the  storage. 

II.   To  find  when  the  net  proceeds  are  due. — Average  the  sales 

alone ^  and  the  result  will  he  the  date  to  he  given  to  the  com,mission 

and  guaranty  ;  then  make  the  sales  the  Cr.  sidcj  and  the  charges 

the  Dr.  side,  and  average  the  entire  account  hy  a  compound  equation, 

Note.  —  In  averaging,  either  the  product  method  or  the  interest  method  may 
be  used. 

EXAMPLES    FOR   PRACTICE. 

1.  Account  sales  of  100  pipes  of  gin,  received  per  ship  Hispan- 
iola,  from  Havana,  on  a|c.  of  Tyler,  Jones  &  Co. 


18  GO 

April 

15 

May 

5 

June 

28 

April 

1 

" 

1 

« 

1 

June 

28 

Sold  32  Pipes,  4160  gal.  @  $1.05,  on  30  days,. 

"    40       ''        6240     "     @     1.02,  cash, 

«•     28      «       3650     "     @     LOO,     ♦'       


6344 
3650 


100 


CHARGES. 


To  Freight  and  Primage, $136.76 

"    Wharfage  and  Cartage 48.54 

"    huty  Bonds,  at  CO  days 3207.07 

"    Storage  from  April  1,  viz. : 

On  32  Pipes,  2  wk.s 64  wks. 

"    40      "       5     "      ...  200     "  - 

"    28      «     13    «      ...  364     « 

100      «      equal  to      628    «@6c 37.68 

«    Commission  on  S13362.80.  at  23^  % 334.07 

"   Guaranty  on  $i368,  at  21^  % 109.20 


3873 


00 
80 
00 

80 


What  are  the  net  proceeds  of  the  ahove  account,  and  when  due  ? 
Ans.  Net  proceeds,  $9489.48 )  due.  May  20,  1860. 

Note. — The  time  for  which  storage  is  charged  on  each  part  of  the  shipment 
is  the  interval,  reduced  to  weeks,  hetween  Apr.  1,  when  the  pipes  were  received 
into  store,  and  the  date  of  sale.     Every  fraction  of  a  wecli  is  reckoned  a  full  week. 

2.  A  commission  merchant  in  Boston  received  into  his  store  on 
May  1,  1859,  1000  bbl.  of  flour,  paying  as  charges  on  the  same 


EQUATION  OF  PAYMENTS. 


363 


day,  freight  $175.48,  cartage  $56.25,  and  cooperage  $8.87.  He 
sold  out  the  shipment  as  follows:  June  3,  200  bbl.  @  $6.25; 
June  30,  850  bbl.  @  $6.50;  July  29,  400  bbl.  @  $3.12 J;  Aug. 
6,  50  bbl.  @  $6.00.  Required  the  net  proceeds,  and  the  date 
when  they  shall  be  accredited  to  the  owner,  allowing  commission 
at  3  J  %,  and  storage  at  2  cents  per  week  per  bbl. 

Ans.  Net  proceeds,  $5614.28 ;  due,  July  10. 


SETTLEMENT   OF  ACCOUNTS    CURRENT. 

©26.  To  find  the  cash  balance  of  an  account  current, 
at  any  given  date. 

/.  Burns  in  account  current  with  Tyler  dh  Co, 


Dr. 

Cr. 

1860. 

1860. 

Feb.       25 

To  Mdse.  on  3  mo. 

360 

75 

March  1 

By  cash  on  acct. 

250 

00 

March  20 

«4            U              it      3        i< 

240 

56 

April  20 

"  accept,  at  30  da. 

300 

00 

April    26 

«       a        «   3     « 

875 

24 

June  12 

"  Sundries, 

375 

00 

June     24 

((      ((       ((  2    « 

235 

26 

«      27 

"  cash  on  acct. 

400 

00 

Required  the  cash  value  of  the  above  account,  July  1,  1860, 
interest  at  6  %. 

OPERATION. 


ur. 

\JT. 

Due. 

Da. 

Items.        Int. 

Cash  val 

Due. 

Da. 

Items.        Int. 

Cash  val. 

May    25 
June   20 
July    26 
Aug.    24 

37 
11 
25 
54 

360.75  +  2.22 
240.56  -f-    .44 
875.24  —  3.65 
235.25—2.12 

362.97 
241.00 
871.59 
233.13 

March  1 
May     20 
June  12 

"      27 

122 
42 
19 
4 

250.00  +  5.08 
300.00  +  2-10 
375.00  +  1.19 
400.00+    .27 

265.08 
302.10 
376.19 
400.27 

1333.64 

1708.69 

^ 

$1708.69— $1333.64  =  $375.05.  Ans. 

Analysis.  For  either  side  of  the  account  we  write  the  dates  at 
which  the  several  items  are  due,  and  the  days  intervening  between 
these  dates  and  the  day  of  settlement,  July  1.  We  then  compute  the 
interest  on  each  item  for  the  corresponding  interval  of  time,  and  add 
it  to  the  item  if  the  maturity  is  before  July  1,  and  subtract  it  from 
the  item  if  the  maturity  is  after  July  1 ;  the  results  must  be  the  cash 
values  of  the  several  items  on  July  1.  Adding  the  two  columns  of 
cash  values,  and  subtracting  the  less  sum  from  the  greater,  we  have 
$375.05.  the  cash  balance  required.     Hence  the 


■ 


864 


PARTNERSHIP. 


E-ULE.  I.  Find  the  number  of  days  intervening  between  each 
maturity  and  the  day  of  settlement. 

II.  Compute  the  interest  on  each  item  for  the  corresponding 
interval  of  time ;  add  the  interest  to  the  item  if  the  maturity  is 
before  the  day  of  settlement,  and  subtract  it  from  the  item  if  the 
maturity  is  after  the  day  of  settlement ;  the  results  will  be  the  cash 
values  of  the  several  items. 

III.  Add  each  column  of  cash  values,  and  the  difference  of  the 
amounts  will  be  the  cash  balance  required. 


EXAMPLES   FOR   PRACTICE. 

1.  Find  the  cash  balance  of  the  following  account  for  June  1, 
1858,  interest  at  6  per  cent.  ? 

Alvan  Parke. 


Dr. 

Cr. 

1858. 

1858. 

Jan.      12 

To  check, 

500 

36 

Jan.       1 

By  bal.  from  old  acct. 

536 

72 

"        26 

a       u 

250 

48 

Feb.       3 

"  cash, 

486 

57 

Feb.      13 

«       u 

400 

00 

March  26 

U          (( 

1250 

78 

March  16 

((       » 

750 

00 

April    20 

((          U 

756 

36 

April     25 

"       " 

200 

00 

May     12 

a      it 

248 

79 

Ans.  ^1196.67. 
2.  What  is  the  cash  balance  of  the  following  account  on  Dec. 
31,  at  7  per  cent.  ? 

James  Hanson. 


Dr. 

Cr. 

1859. 

1859. 

8«pt.    3 

To  Sandriea, 

478 

36 

Sepk.  17 

By  Sundries, 

96 

54 

Oct.      2 

"  Mdse.  on  3  mo. 

256 

37 

«      20 

"  cash  on  acct. 

200 

00 

"     21 

"      "       "  3    " 

375 

.   26 

f  oa 

Oct.      S 

(i      a            u 

325 

00 

Nov.   12 

«      «       «  3    « 

80 

Nov.   17 

«      ((            (( 

50 

00 

Dec.    15 

"  Sundries, 

148 

1   it 

Dec.    27 

«      «            « 

84 

00 

PAETNERSHIP. 

037*  Partnership  is  a  relation  established  between  two  or 
more  persons  in  trade,  by  which  they  agree  to  share  the  profits 
and  losses  of  business  according  to  the  amount  of  capital  furnished 
by  each,  and  the  time  it  is  employed. 

6^8.   The  Partners  are  the  individuals  thus  associated. 

Note.  — The  terms  Capital  or  Stock,  Dividendj  and  Assessment,  have  the  same 
?ignificafcion  in  Partnership  as  in  Stocks. 


PARTNERSHIP.  365 

CASE  I. 

6S9.  To  find  each  partner's  share  of  the  profit  or 
loss,  when  their  capital  is  employed  for  equal  periods 
of  time. 

1.  A  and  B  engage  in  trade;  A  furnishes  $500,  and  B  $700 as 
capital;  they  gain  $96;  what  is  each  man's  share  ? 

OPERATION.  Analysis.  The  whole 

$  500  amount  of  capital  em- 

$  700  ployed  is  $500  +  $700 

$1200,  whole  stock.  =$1200 ;  hence,  A  fur- 

^5  0  0  ^   5^^  A^s  part  of  the  stock.  ^^«^?^«  t¥A  =t2^{  the 

\^Q0_ 7*   JVa     ic     u     u     u  capital,  and  B  furnishes 

|96  X  A  =  $40  A's  share  of  the  gain.      t^Vo  =  h  of  t^e  capi- 

$90  X  Ss  =  $56,  B's     '^     ''     "     ''         *^^-     ^""^  '^^'^  ^^'^ 

man's  share  of  the  pro- 
fit or  loss  will  have  the  same  ratio  to  the  whole  profit  or  loss  as  his 
part  of  the  capital  has  to  the  whole  capital,  A  will  have  /g  of  the 
$9G,  and  B  y\  of  the  $96,  for  their  respective  shares  of  the  profits. 

We  may  also  regard  the  whole  capital  as  the  first  cause^  and  each 
man's  share  of  the  capital  as  the  second  cause,  the  whole  profit  or  loss 
as  the  first  effect,  and  each  man's  share  of  the  profit  or  loss  as  the 
second  effect,  and  solve  by  proportion  thus : 

1st  cause.  2d  cause.    1st  efifect.      2d  effect. 

$1200     :  :  $500  =  $96     :     (  ?  )  =  $40,  A's  gain, 
$1200     :     $700  =  $96     :     (?)  =  $56,  B's     " 
Hence  we  have  the  following 

BuLE.  Multiple/  the  whole  projli  or  loss  hy  the  ratio  of  the 
whole  capital  to  each  man's  share  of  the  capital.    Or, 

The  whole  capital  is  to  each  man^s  share  of  the  capital  as  the 
whole  profit  or  loss  is  to  each  man's  share  of  the  profit  or  loss» 

EXAMPLES   FOR   PRACTICE. 

1.  Three  men  engage  in  trade;  A  puts  in  $6470,  B  $3780,  and 
C  $9860,  and  they  gain  $7890.  What  is  each  partner's  share  of  the 
profit?    Ans.  A%  $2538.453 ;  B's,  $1483.053;  C's,  $3868^.493. 

2.  B  and  C  buy  pork  to  the  amount  of  $1847.50,  of  which  B 
pays  $739,  and  C  the  remainder.  They  gain  $375 ;  what  is  each" 
one's  share  of  the  gain  ?  ' 

31* 


366  ,  PARTNERSHIP. 

3.  A,  B,  and  C  form  a  company  for  the  manufacture  of  woolen 
cloths.  A  puts  in  $10000,  B  $12800,  and  C  $3200.  C  is  allowed 
$1500  a  year  for  personal  attention  to  the  business;  their  ex- 
penses for  labor,  clerk  hire,  and  other  incidentals  for  1  year  are 
$3400,  and  their  receipts  auring  the  same  time  are  $9400.  What 
is  A^s,  B's,  and  C's  income  respectively  from  the  business  ? 

4.  Four  persons  rent  a  farm  of  115  A.  32  P.  at  $3.75  an  acre. 
A  puts  on  144,  B  160,  C  192,  and  D  324  sheep;  how  much  rent 
ought  each  to  pay  ? 

5.  Three  persons  gain  $2640,  of  which  B  is  to  have  $6  as  often 
as  C  $4,  and  as  often  as  D  $2 ;  how  much  is  each  one's  share  ? 

6.  Six  persons  are  to  share  among  them  $6300 ;  A  is  to  have 
^  of  it,  B  i,  C  |,  D  is  to  have  as  much  as  A  and  C  together,  and 
the  remainder  is  to  be  divided  between  E  and  F  in  the  ratio  of 
3  to  5.     How  much  does  each  receive  ? 

Arts.  A,  $900;  B,  $1260;  C,  $1400; 
D,  $2300;  E,  $165;  F,  $275. 

7.  Two  persons  find  a  watch  worth  $90,  and  agree  to  divide  the 

value  of  it  in  the  ratio  of  |  to  | ;  how  much  is  each  one's  share  ? 

Note. — If  the  fractions  be  reduced  to  a  common  denominator,  they  will  be  to 
each  other  as  their  numerators,  (418,  III). 

8.  A  father  divides  his  estate  worth  $5463.80  between  his  two 
sons  giving  the  elder  J  more  than  the  younger ;  how  much  is  each 
son's  share  J*  Ans.  Elder,  $2892.60;  younger,  $2571.20. 

9.  Three  men  trade  in  company.  A  furnishes  $8000,  and  B 
$12000  Their  gam  is  $1680,  of  which  C's  share  is  $840;  required, 
C's  stock,  and  A's  and  B's  gain.  Ans,  C's  stock,  $20,000. 

10.  Four  persons  engage  in  the  lumber  trade,  and  invest  jointly 
$22500;  at  the  expiration  of  a  certain  time,  A's  share  of  the 
gain  is  $2000,  B's  $2800.75,  C's  $1685.25,  and  D's  $1014;  how 
much  capital  did  each  put  in  ?  Ans.  I>  put  in  $3042. 

11.  A  legacy  of  $30,000  was  left  to  four  heirs  in  the  propor- 
tion of  ^,  I,  I,  and  5,  respectively;  how  much  was  the  share  of 
each? 

12.  Three  men  purchase  a  piece  of  land  for  $1200,  of  which 
sum  C  pays  $500.     They  seli  it  so  as  to  gain  a  certain  sum^  of 


PARTNERSHIP.  367 

which  A  takes  $71.27,  and  B  $142.54;  how  much  do  A  and  B 
pay,  and  what  is  C's  share  of  the  gain  ?    Ans,  C's  gain,  $152.72 1. 

13.  Three  persons  enter  into  partnership  for  the  manufacture 
of  coal  oil,  with  a  joint  capital  of  $18840.  A  puts  in  $3  as  often 
as  B  puts  in  $5,  and  as  often  as  C  puts  in  $7.  Their  annual  gain 
is  equal  to  C's  stock;  how  much  is  each  partner's  gain? 

14.  Ay  B,  and  C  are  employed  to  do  a  piece  of  work  for  $26.45. 
A  and  B  together  are  supposed  to  do  |  of  the  work,  A  and  C  -j^^, 
and  B  and  C  ^^,  and  are  paid  proportionally;  how  much  must 
each  receive?  Ans,  A,  $11.50;  B,  $575;  C,  $9.20. 

CASE  ir. 

030.  To  find  each  partner's  share  of  the  profit  or  loss 
when  their  capital  is  employed  for  unequal  periods  of 
time. 

It  is  evident  that  the  respective  shares  of  profit  and  loss  will 
depend  equally  upon  two  conditions,  viz.:  the  amount  of  capital 
invested  by  each,  and  the  time  it  is  employed.  Hence  they  will 
be  proportional  to  the  products  of  these  two  elements. 

1.  Two  men  form  a  partnership ;  A  puts  in  $320  for  5  months, 
and  B  $400  for  6  months.  They  lose  $140 ;  what  is  each  man's 
share  of  the  loss  ? 

OPERATION. 

$320  X  5  =  $1600,  A^s  capital  for  1  mo. 
$400  X  6  =  $2400,  B's       ''       ''       " 

$4000,  entire  '^        '^       '^ 
$lg-Q-0  =  |,  A's  share  in  the  partnership 

$140  X  I  =  $56,  A^«  loss- 
$140  X  I  =  $84,  B's  loss. 

Analysis  The  use  of  $320  for  5  months  is  the  same  as  the  use  of 
5  times  $320,  or  $1G00,  for  1  month  ;  and  the  use  of  $400  for  6  months 
is  the  same  as  the  use  of  6  times  $400,  or  $2400,  for  1  month ;  hence 
the  use  ot  the  entire  capital  is  the  same  as  the  use  of  $1000  +  $2400 
==r  $4000  for  1  month.  A^s  interest  in  the  partnership  is  therefore 
\l\\  =  §,  and  he  will  suffer  §  of  the  loss,  or  $140  X  f  —  $56  .•  and 


368^  PARTNERSHIP. 

B's  interest  in  the  partnership  is  f  ^  J§  =  |,  and  he  will  suffer  |  of  the 

loss,  or  $140  X  §-=$84. 

We  may  also  solve  by  proportion,  the  causes  being  compounded  of 

the  two  elements,  capital  and  time  •  thus : 

84000  :  81600  =  $140  :  (?)  =  $56,  A^s  loss, 
$4000  :  $2400  =  $140  :  (?)  =  $84,  B's  loss. 

Hence  the  following 

Rule.  3/wZ^/jjZ^  each  maiis  cajntal  hy  tlie  time  it  is  employed 
in  tradey  and  add  tlie  products.  Then  multiply  the  entire  profit  or 
loss  hy  the  ratio  of  eacji  product  to  the  sum  of  the  products ;  the 
results  will  he  the  respective  shares  of  profit  or  loss  of  each  part- 
ner.    Or, 

Midtiply  each  man^s  capital  hy  the  time  it  is  emptloycd  in  trade, 
and  regard  each  product  as  his  capital,  and  the  sum  of  the  p7'o^ 
ducts  as  the  entire  cap>ital,  and  solve  ly  proportion,  as  in  Case  I. 

EXAMPLES    FOR   PRACTICE. 

1.  A,  11,  and  C  enter  into  partnership.  A  puts  in  $357  for  5 
months,  B  $371  for  7  months,  and  C  $154  for  11  months,  and  they 
gain  $347.20;  how  much  is  each  one's  share? 

Ans.  A^s  $102;  B's  $148.40;  C's  $96.80. 

2.  Three  men  hire  a  pasture  for  $55.50.  A  put  in  5  cows,  12 
weeks;  B,  4  cows,  10  weeks;  and  C,  6  cows,  8  weeks;  how  much 
ought  eacii  to  pay?  Ans.  A  $22.50;  B  $15;  C  $18. 

3.  B  commenced  business  with  a  capital  of  $15000.  Three 
months  afterward  C  entered  into  partnership  with,  him,  and  .put 
in  125  acres  of  land.  At  the  close  of  the  year  their  profits  were 
$4500,  of  which  C  was  entitled  to  $1800 ;  what  was  the  value  of 
the  land  per  acre  ? 

4.  A  and  B  engaged  in  trade.  A  put  in  $4200  at  first,  and  9 
months  afterward  $200  more.  B  put  in  at  first  $1500,  and  at  the 
end  of  6  months  took  out  $500.  At  the  end  of  16  months  their 
gain  was  $772.20 ;  how  much  is  the  share  of  each  ? 

5.  Four  companies  of  men  worked  on  a  railroad.  In  the  first 
company  there  were  30  men  who  worked  12  days,  9  hours  a  day; 
in  the  second,  there  were  32  men  who  worked  15  days,  10  hours 


PARTNERSHIP. 


369 


a  day;  in  the  Ihird,  there  were  28  men  who  worked  18  days,  11 
hours  a  day;  and  in  the  fourth,  there  were  20  men  who  worked 
15  days,  12  hours  a  day.  The  entire  amount  paid  to  all  the  com- 
panies was  $1500;  how  much  were  the  wages  of  each  company? 

6.  A  and  B  are  partners.  A's  capital  is  to  B's  as  5  to  8 ;  at 
the  end  of  4  months  A  withdraws  J  of  his  capital,  and  B  |  of  his ; 
at  the  end  of  the  year  their  whole  gain  is  $4000 ;  how  much  be- 
longs to  each  ?  Ans.  A,  $17141 ;  B,  $2285^; 

7.  B,  C,  and  D  form  a  manufacturing  company,  with  capitals 
of  $15800,  $25000,  and  $30000  respectively.  After  4  months  B 
draws  out  $1200,  and  in  2  months  more  he  draws  out  $1500  more, 
and  4  months  afterward  puts  in  $1000.  C  draws  out  $2000  at 
the  end  of  6  months,  and  $1500  more  4  months  afterward,  and  a 
month  later  puts  in  $800.  D  puts  in  $1800  at  the  end  of  7 
months,  and  3  months  after  draws  out  $5000.  If  their  gain  at 
the  end  of  18  months  be  $15000,  how  much  should  each  receive? 

Ans.  B,  $3228.07;  C,  $5258.15;  D,  $6513.78. 

8.  The  joint  stock  of  a  company  was  $5400,  which  was  doubled 
at  the  end  of  the  year.  A  put  i  for  J  of  a  year,  B  |  for  J  a  year, 
and  C  the  remainder  for  one  year.  How  much  is  each  one^s  share 
of  the  entire  stock  at  the  end  of  the  year  ? 

9.  Three  men  engage  in  merchandising.  A^s  money  was  in 
10  months,  for  which  he  received  $456  of  the  profits ;  B's  was  in 
8  months,  for  whioh  he  received  $343.20  of  the  profits;  and  C's 
was  in  12  months,  for  which  he  received  $750  of  the  profits.  Their 
whole  capital  invested  was  $14345 ;  how  much  was  the  capital  of 
each?  An$,  A's,  $4332;  B's,  $4075.50;  C's,'  $5937.50. 

10.  Three  men  take  an  interest  in  a  coal  mine.  B  invests  his 
capital  for  4  months,  and  claims  -j'^  of  the  profits;  C's  capital  is  in 
8  months ;  and  D  invests  $6000  for  6  months,  and  claims  |  of  the 
profits ;  how  much  did  B  and  C  put  in  ? 

11.  A,  B,  and  C  engage  in  manufacturing  shoes.  A  puts  in 
$1920  for  6  months;  B,  a  sum  not  specified  for  12  months;  and 
C,  $1280  for  a  time  not  specified.  A  received  $2400  for  his  stock 
and  profits,  B  $4800  for  his,  and  C  $2080  for  his.  Required, 
B's  stock,  and  C's  time  ? 

Y 

\ 


870  ALLIGATION. 


ALLIGATION. 

631.  Alligation  treats  of  mixing  or  compounding  two  or 
more  ingredients  of  different  values  or  qualities. 

63S.  The  Mean  Price  or  ftnality  is  the  average  price  or 
quality  of  the  ingredients,  or  the  price  or  quality  of  a  unit  of  the 
mixture. 

CASE   I. 

633.  To  find  the  mean  price  or  quality  of  a  mixture, 
when  the  quantity  and  price  of  the  several  ingredients 
are  given. 

Note. — The  process  of  finding  the  mean  or  average  price  of  several  ingredi- 
ents is  called  AUxyation  Medial. 

1.  A  produce  dealer  mixed  together  84  bushels  of  oats  worth 
$.30  a  bushel,  60  bushels  of  oats  worth  $.38  a  bushel,  and  56 
bushels  of  oats  worth  $  40  a  bushel ;  required,  the  mean  price. 

OPERATION.  Analysis.     The  worth  of  84 

$.30  X  84  =  $25.20  bushels  @  $.30  is  $25.20,  of 

.38  X  60  =    22.80  60  bushels  @  $.38  is  $22.80, 

.40  X  56  =    22.40  and  of  56  bushels  @  $.40  is 

200     ^  ^70  40  $22.40 ;  and  we  have  in  the 

;; whole  compound  84  +  60  +  56 

$.3520,  Ans,  —200  bushels,  worth  $25.20+ 

$22.80  +  $22.40  =  $70.40.     One  bushel  of  the  mixture  is  therefore 
worth  $70.40  -h  200  =  $.352.     Hence  the  following 

Rule.  Find  the  entire  cost  or  value  of  the  xngredientSy  and 
divide  it  hy  the  sum  of  the  simples. 

EXAMPLES   FOR   PRACTICE. 

1.  A  grocer  mixed  4  lb.  of  tea  at  $.60  with  3  lb.  at  $.70,  1  lb. 
at  $1.10,  and  2  lb.  at  $1.20;  how  much  is  1  lb.  of  the  mixture 
worth?  -^/is.  $.80. 

2.  A  dealer  in  liquors  would  mix  14  gal.  of  water  with  12  gal. 
of  wine  at  $.75,  24  gal.  at  $.90,  and  16  gal.  at  $1.10;  how  muck 
is  a  gallon  of  the  mixture  worth  ?  Ans.  $.73^^-5. 


ALLIGATION.  371 

3.  If  3  lb.  6  oz.  of  gold  23  carats  fine  be  compounded  with 
^4  lb.  8  oz.  21  carats,  3  lb.  9  oz.  20  carats,  and  2  lb.  2  oz.  of  alloy, 

what  is  the  fineness  of  the  composition  ?  Ans,  18  carats. 

4.  A  grain  dealer  mixes  15  bu.  of  wheat,  at  $1.20  with  5  bu. 
at  $1.10,  5  bu.  at  S.90,  and  10  bu.  at  $.70 ;  what  will  be  his  gain 
per  bushel  if  he  sell  the  compound  at  $1.25. 

5.  A  merchant  sold  17  lb.  of  sugar  at  5  cts.  a  pound,  51  lb.  at 
8  cts.,  68  lb.  at  10  cts.,  17  lb.  at  12  cts.,  and  thereby  gained  on 
the  whole  33 J  per  cent;  how  much  was  the  average  cost  per 
pound  ? 

6.  A  drover  bought  42  sheep  at  $2.70  per  head,  48  at  $2.85, 
and  65  at  $3.24 ;  at  what  average  price  per  head  must  he  sell 
them  to  gain  20  per  cent.?  Ans,  $3.567^ f. 

7.  A  surveyor  took  10  sets  of  observations  with  an  instrument, 
for  the  measurement  of  an  angle,  with  the  following  results :  1st, 
36°  17'  25.4";  2d,  36°  17'  24.5";  3d,  36°  17'  27.8";  4th,  36°  17' 
26.9";  5th,  36°  17'  25.4";  6th,  36°  17'  24.7";  7th,  36°  17'  24.2"; 
8th,  36°  17'  26.3";  9th,  36°  17'  25.8";  10th,  36°  17'  26.7".  What 
is  the  average  of  these  measurements  ?       Ans,  36°  17'  25.77''. 

8.  Three  trials  were  made  with  chronometers  to  determine  the 
difi'erence  of  time  between  two  places;  the  first  trial  gave  37  min. 
54.16  sec,  the  second  37  min.  55.56  sec,  and  the  third  37  min. 
54.82  sec  Owing  to  the  favorable  conditions  of  the  third  trjal, 
it  is  entitled  to  twice  the  degree  of  reliance  to  be  placed  upon 
either  of  the  others ;  what  should  be  taken  as  the  difference  of 
longitude  between  the  two  places,  according  to  these  observations? 

Ans,  9°  28' 42.6". 

CASE    IL 

634.  To  find  the  proportional  quantity  to  be  used  of 
each  ingredient,  when  the  mean  price  and  the  prices  of 
the  several  simples  are  given. 

Note. — The  process  of  finding  the  quantities  to  be  used  in  any  required  mix- 
ture is  commonly  called  Alliyation  Alternate. 

1.  A  farmer  would  mix  oats  worth  3  shillings  a  bushel  with 
peas  worth  8  shillings  a  bushel,  to  make  a  compound  worth  5  shil- 
lings a  bushel ;  what  quantities  of  each  may  he  take  ? 


372 


ALLIGATION. 


oPERiTiON.  Analysis.     If  a  mixture,  in  any  pro- 

^  o  I  ,  I  Q  \  portions,  of  oats   worth  3  shillings    a 

5  -<  o     ?    9  [■  Ans.  bushel  and  peas  worth  8  shillings,  be 

^      ^        ^  priced  at  5   shillings,  there  will  be  a 

gain  on  the  oats,  the  ingredient  worth  less  than  the  mean  price,  and 
a  loss  on  the  peas,  the  ingredient  worth  moi^e  than  the  mean  price ; 
and  if  we  take  such  quantities  of  each  that  the  gain  and  loss  shall 
each  be  1  shilling,  the  unit  of  value,  the  result  will  be  the  required 
mixture.  By  selling  1  bushel  of  oats  worth  3  shillings  for  5  shil- 
lings, there  will  be  a  gain  of  5  —  3  =  2  shillings,  and  to  gain  1  shil- 
ling would  require  J  of  a  bushel ;  hence  we  place  i  opposite  the  3.  By 
selling  1  bushel  of  peas  worth  8  shillings  for  5  shillings,  there  will 
be  a  loss  of  8  —  5  =  3  shillings,  and  to  lose  1  shilling  will  require  J 
of  a  bushel ;  hence  we  write  J  opposite  the  8.  Therefore,  }  bushel  of 
oats  to  :J  of  a  bushel  of  peas  are  the  propoy^tional  quantities  for  the 
required  mixture.  It  is  evident  that  the  gain  and  loss  will  be  equal, 
if  we  take  any  number  of  times  these  proportional  terms  for  the  mix- 
ture. We  may  therefore  multiply  the  fractions  J  and  J  by  6,  the  least 
common  multiple  of  their  denominators,  and  obtain  the  integers  3 
and  2  for  the  proportional  terms  (418,111);  that  is,  we  may  take,  for  the 
mixture,  3  bushels  of  oats  to  2  bushels  of  peas. 

2.  What  relative  quantities  of  sugar  at  7  cents,  8  cents,  11  cents, 
and  14  cents  per  pound,  will  produce  a  mixture  worth  10  cents 
per  pound  ? 

OPERATION.  Analysis.      To    preserve    the 

equality  of  gains  and  losses,  we 
must  compare  two  prices  or  sim- 
ples, one  greater  and  one  less  than 
10  ^  -n  -,  o     r»  the  mean  rate,  and  treat  each  pair 

or  couplet  as  a  separate  example. 
Thus,  comparing  the  simples  whose 
prices  are  7  cents  and  14  cents,  we 
find  that,  to  gain  1  cent,  J  of  a 
pound  at  7  cents  must  be  taken, 
and,  to  lose  1  cent,  J  of  a  pound  at 
14  cents  must  be  taken  ;  and  com- 
paring the  simples  the  prices  of 
which  are  8  cents  and  11  cents,  we 
find  that  J  pound  at  8  cents  must  be  taken  to  gain  1  cent,  and  1  pound 
At  11  cents  must  be  taken  to  lose  1  cent.    These  proportional  terms  are 


11 
14 


1 

2 

3 

4 

5 

T 

4 

4 

i 

1 

1 

1 

2 

2 

i 

3 

3 

10^ 


1 

2 

3 

7 

T 

1 

8 

J 

4 

11 

1 

3 

14 

} 

2 

ALLIGATION.  373 

"written  in  columns  1  and  2  We  now  reduce  these  couplets  separately 
to  integers,  as  in  the  last  example,  writing  the  results  in  columns  3 
and  4 ;  and  arranging  all  the  terms  in  column  5,  we  have  4,  1,  2,  and 
3  for  the  proportional  quantities  required.  If  we  compare  the  prices 
7  and  11  fcr  the  first  couplet,  and  the  prices  8  and  14  for  the  second 
couplet,  as  in  the  second  operation,  we  shall  obtain  1,  4,  3.,.  and  2  fo^ 
the  proportional  terms. 

It  will  be  seen  that  in  comparing  the  simples  of  any  couplet,  one 
of  which  is  greater  and  the  other  less  than  the  mean  rate,  the  pro- 
portional number  finally  obtained  for  either  term  is  the  difi'erence 
between  the  mean  rate  and  the  other  term.  Thus,  in  comparing  7 
and  14,  the  proportional  number  corresponding  to  the  former  simple 
is  4,  which  is  the  difi'erence  between  14  and  the  mean  rate  10 ;  and 
the  proportional  number  corresponding  to  the  latter  simple  is  3, 
which  is  the  difi'erence  between  7  and  the  mean  rate.  The  same  is 
true  of  every  other  couplet.  Hence,  when  the  simples  and  the  mean 
rate  are  integers,  the  intermediate  steps  taken  to  obtain  the  final  pro- 
portional numbers  as  in  columns  1,  2,  3,  and  4,  may  be  omitted,  and 
the  same  results  readily  found  by  taking  the  difi'erence  between  each 
simple  and  the  mean  rate,  and  placing  it  opposite  the  one  with  which 
it  is  compared. 

From  these  examples  and  analyses  we  derive  the  following 
EuLE.     I.    Write   the  prices   or  qualities  of  the  several  ingre- 
dients in  a  column^  and  the  mean  price  or  quality  at  the  left. 

II.  Consider  any  two  prices^  one  of  which  is  less  and  the  other 
greater  than  the  mean  ratCy  as  forming  a  couplet ;  find  the  differ- 
ence  between  each  of  these  prices  and  the  mean  rate,  and  write  the 
reciprocal  of  each  difference  opposite  the  given  price  m  the  couplet^ 
as  one  of  the  proportional  terms.  In  like  manner  form  the  couplets ^ 
till  all  the  prices  have  been  employed,  writing  each  pair  of  propor- 
tional terms  in  a  separate  column. 

III.  If  the  proportional  terms  thus  obtained  are  fractional,  mul- 

tijyly  each  pair  by  the  least  common  multiple  of  their  denominators, 

and  carry  these  integral  products  to  a  single  column^  observing  to 

add  any  two  or  more  that  stand  in  the  same  horizontal  line;  the 

final  results  will  be  the  proportional  quantities  required. 

NoTKS. — 1.  If  the  numbers  in  any  couplet  or  column  have  a  common  factor, 
it  may  be  rejected. 

32      \ 


374  ALLIGATION. 

2.  We  may  also  multiply  the  numbers  in  any  couplet  or  column' by  any  mul- 
tiplier we  choose,  without  affecting  the  equality  of  the  gains  and  losses,  and 
thus  obtain  an  indefinite  number  of  results,  any  one  of  which  being  taken  will 
give  a  correct  final  result, 

EXAMPLES    FOE   PRACTICE. 

1.  What  quantities  of  flour  worth  $5i,  $6,  and  $71  per  barrel, 
must  be  sold,  to  realize  an  average  price  of  $6i  per  barrel? 

OPERATION.  Analysis.       Comparing    the 

r  ^1     4        14  A  first  price  with  the  third,  we  ob- 

0J.  J  g  4  j         12    12  ^^^^  *^®  couplet  J  to  t;  and  com- 

(74ffj2      2      4  paring  the  second  price  with  the 

third,  we  obtain  the  couplet  4  to 
•J.  Reducing  these  proportional  terms  to  integers,  we  find  that  we 
may  take  4  barrels  of  the  first  kind  with  2  of  the  third,  and  12  of  the 
second  kjnd  with  2  of  the  third ;  and  these  two  combinations  taken 
together  give  4  of  the  first  kind,  12  of  the  second,  and  4  of  the  third. 

2.  How  much  sugar  worth  5  cts.,  7  cts.,  12  cts.,  and  13  cts.  per 
pound,  will  form  a  mixture  worth  10  cts.  per  pound? 

3  lb.  of  each  of  the  first  and  third  kinds,  2  lb. 
of  the  second,  and  5  lb.  of  the  fourth. 
8.  How  can  wine  worth  $.60  $.90  and  $1.15  per  gallon  be  mixed 
with  water  so  as  to  form  a  mixture  worth  $.75  a  gallon  ? 

J       (By  taking  3  gal.  of  each  of  the  first  two  kinds  of 
1    wine,  15  gal.  of  the  third,  and  8  gal.  of  water. 

4.  A  farmer  has  3  pieces  of  land  worth  $40,  $60,  and  $80  an 
acre  respectively.  How  many  acres  must  he  sell  from  the  dif- 
ferent tracts,  to  realize  an  average  price  of  $62.50  an  acre? 

5.  How  much  wine  worth  $.60,  $.50,  $.42,  $.38,  and  $.30  per 
pint,  will  make  a  mixture  worth  $.45  a  pint  ? 

6.  What  relative  quantities  of  silver  |  pure,  |  pure,  and  ^^^ 
pure,  will  make  a  mixture  |  pure  ? 

Ans.   3  lb.  J  pure,  3  lb.  |  pure,  and  20  lb.  j%  pura. 

CASE   III. 

63«>.  "When  two  or  more  of  the  quantities  are  es- 
quired to  be  in  a  certain  proportion. 

1-  A  farmer  having  oats  worth  $.30,  corn  worth  $.60,  and  wheat 


ins.  < 


ALLIGATION. 


375 


r   30 

50 )    60 

(no 


1 

2 

3 

4 

5 

6 

^ 

7u 

1 

3 

6 

7 

t'd 

2 

2 

e^i 

1 

2 

2 

worth  $1.10  per  bushel,  desires  to  form  a  mixture  worth  §.50  per 
bushel,  which  shall  contain  equal  parts  of  corn  and  wheat ;  in 
what  proportion  shall  the  ingredients  be  taken  ? 

OPERATION.  Analysis.     We  first  obtain 

the  proportional  terms  in  col- 
umns 3  and  4,  by  Case  II. 
Now,  it  is  evident  that  the  loss 
and  gain  will  be  equal  if  we 
take  each  couplet,  or  any  mul- 
tiple of  each,  alone ;  or  both 
couplets,  or  any  multiples  of  both,  together.  Multiplying  the  terms 
in  column  4  by  2,  we  obtain  the  terms  in  column  5  ;  and  adding  the 
terms  in  columns  3  and  5,  we  obtain  the  terms  in  column  6 ;  that  is, 
the  farmer  takes  7  bushels  of  oats  to  2  of  corn  and  2  of  wheat,  which 
is  the  required  proportion.     Hence  the  following 

IlULE.  I.  Compare  the  given  prices,  and  obtain  the  proportional 
terms  by  couplets,  as  in  Case  IL 

II.  Reduce  the  couplets  to  higher  or  lower  terms,  as  may  be  re- 
quired;  then  select  the  columns  at  pleasure,  and  combine  them  by 
adding  the  terms  in  the  same  horizontal  Ihve,  till  a  set  of  pro- 
portional terms  is  obtained,  answeri7ig  the  required  conditions. 

EXAMPLES   FOR   PRACTICE. 

1.  A  grocer  has  four  kinds  of  molasses,  worth  $.25,  $.50,  $.62, 
and  $.70  per  gallon,  respectively ;  in  what  proportions  may  he  mix 
the  four  kinds,  to  obtain  a  compound  worth  $.58  per  gallon,  using 
equal  parts  of  the  first  two  kinds  ?  Ans.  4,  4,  8  and  11. 

2.  In  what  proportions  may  we  take  sugars  at  7  cts.,  8  cts.,  13 
cts.,  and  15  cts.,  to  form  a  compound  worth  10  cts.  per  pound,  using 
equal  parts  of  the  first  three  kinds  ?  Ans.  5,  5,  5  and  2. 

3  A  miller  has  oats  at  30  cts.,  corn  at  50  cts.,  and  wheat  at 
100  cts.  per  bushel.  He  desires  to  form  two  mixtures,  each  worth 
70  cts.  per  bushel.  In  the  first  he  would  have  equal  parts  of  oats 
and  corn,  and  in  the  second,  equal  parts  of  corn  and  wheat;  what 
must  be  the  proportional  terms  for  each  mixture  ? 

J,        (  For  the  first  mixture,  1,  1  and  2. 
1  For  the  second  mixture,  1,  4  and  4. 


376 


ALLIGATION. 


OPERATION. 


r28  1 
58  ]  441 ' 


3'tT 

1 

T4 

7 
5 

140 
100 

3*5 

.'s 

6     2 

8 

160 

CASE    IV. 

636.  Wlien  the  quantity  of  one  of  the  simples  is 
limited. 

1.  A  miller  has  oats  worth  ^.28,  corn  worth  8.44,  and  barley 
worth  $.90  per  bushel.  He  wishes  to  form  a  mixture  worth  $.58 
per  bushel,  and  containing  100  bushels  of  corn.  How  many 
bushels  of  oats  and  barley  may  he  take  ? 

Analysis.  By  Case 
II,  we  find  the  pro- 
portional quantities 
to  be  7  bushels  of 
oats  to  5  of  corn  and 
8  of  barley.  But,  as  100  bushels  of  corn,  instead  of  5,  are  required, 
we  must  take  '  J*^  =20  times  each  of  the  other  ingredients,  in  order 
that  the  gain  and  loss  may  be  equal ;  and  we  shall  therefore  have 
7  X  20  =  140  bushels  o.  oats,  and  8  X  20  =  160  bushels  of  barley. 
Hence  the  following 

HuLE.  Find  the  proportional  quantities  by  Case  II  or  Case 
III.  Divide  ihe^  given  quantity  hy  the  proportional  quantity  of 
this  ingredient  J  and  multiply  each  of  the  other  proportional  quan- 
tities hy  the  quotient  thus  obtained, 

EXAMPLES   FOR   PRACTICE. 

1.  A  dairyman  bought  10  cows  at  $20  a  head ;  how  many  must 
he  buy  at  $16,  $18,  and  $24  a  head,  so  that  the  whole  may  cost 
him  an  average  price  of  $22  a  head  ? 

Ans,  10  at  $16,  10  at  $18,  and  60  at  $24. 

2.  Bought  12  yards  of  cloth  for  $15  •  how  many  yards  must  I 
buy  at  $lf ,  and  $i  a  yard,  that  the  average  price  of  the  whole 
may  be  $1^?  Ans,  12  yards  at  $1J  and  16  yards  at  $|. 

3.  How  much  water  will  dilute  9  gal.  2  qt.  1  pt.  of  alcohol  96 
per  cent,  strong  to  84  per  cent.  ?        -       Ans.  1  gal.  1  qt.  1  pt. 

4.  A  grocer  mixed  teas  worth  $.30,  $.55,  and  $.70  per  pound 
respectively,  forming  a  mixture  worth  $.45  per  pound,  having  equal 
parts  of  the  first  two  kinds,  and  12  lbs.  of  the  third  kind;  hoT? 
many  pounds  of  each  of  the  first  two  kinds  did  he  take  ? 


ALLIGATION. 


377 


OPERATION. 

$.48  X  18  =  §  8.64 

.52  X    8  =  4.16 

.85  X  J  =  3.40 

30    )  SI  6.20 

Mean  price  of  the  )  ^      ^j_ 
given  simples      J 


( 

<■   54 

1114. 

2  6  30 

84. 

108 

^■4        5 

5  25 

{ 

Ll44 

gV 

1  1     5 

CASE    V. 

637.  When  the  quantities  of  two  or  more  of  the  in- 
gredients are  limited. 

1.  How  many  bushels  of  rye  at  $1.08,  and  of  wheat  at  $1.44, 
must  be  mixed  with  18  bushels  of  oats  at  $.48,  8  bushels  of  corn 
at  $52,  and  4  bushels  of  barley  at  $.85,  that  the  mixture  may  be 
worth  $.84  per  bushel  ? 

Analysis.  Of  the  given 
quantities  there  are  18  -{- 
8  +  4  =  30  bushels,  whose 
mean  or  average  price  we 
find  by  Case  I  to  be  $.54. 
We  are  therefore  required  to 
mix  30  bushels  of  grain 
worth  $.54  per  bushel,  with 
rye  at  $1.08,  and  wheat  at 
$1.44,  to  make  a  compound 
worth  $.84  per  bushel.  Pro- 
ceeding as  in  Case  IV,  we 
find  there  will  be  required  25  bushels  of  rye,  and  5  bushels  of  wheat. 
Hence  the  following 

Rule.  Consider  those  ingredients  whose  quantities  and  prices 
are  given  as  forming  a  mixture  y  and  find  their  mean  price  hy  Case 
I;  then  consider  this  mixture  as  a  single  ingredient  whose  quantify 
and  price  are  known,  andfitid  the  quantities  of  the  other  ingredients 
hy  Case  IV, 

EXAMPLES    FOR   PRACTICE. 

1.  A  gentleman  bought  7  yards  of  cloth  @  $2.20,  and  7  yards 
@  $2  j  how  much  must  he  buy  @  $1.60,  and  @  $1.75  that  the 
average  price  of  the  whole  may  be  $1.80  ? 

2.  How  much  wine,  at  $1.75  a  gallon,  must  be  added  to  60  gal- 
lons at  $1.14,  and  30  gallons  at  $1.26  a  gallon,  so  that  the  mixture 
may  be  worth  $1.57  a  gallon  ?  Ans.  195  gallons. 

3.  A  farmer  has  40  bushels  of  wheat  worth  $2  a  bushel,  and 
70  bushels  of  corn  worth  $J  a  bushel.  How  many  oats  worth  $} 
a  bushel  must  he  mix  with  the  wheat  and  corn,  to  make  the  mix- 
ture worth  $1  a  bushel  ?  Ans.  6|  bushels. 

32  * 


378 


ALLIGATION. 


CASE  VI. 

638.  When  the  quantity  of  the  whole  compound  is 
limited. 

1.  A  tradesman  has  three  kinds  of  tea  rated  at  $.30,  $.45,  and 
$.60  per  pound,  respectively;  what  quantities  of  each  should  he 
take  to  form  a  mixture  of  72  pounds,  worth  $.40  per  pound? 

OPERATION.  Analysis.    By  Case  II, 
1       o       o     ^     r      r»               we  find  the   proportional 
1      2      3    4    5     6  ^      /         XI. 
quantities    to     lorm    tne 

(SO    TU    tV     2    13    36  mixture    to  be   3  lb.   at 

40 }  45  I  2    2    24  $.30,  2  lb.  at  $.45,  and 

(OO    ^jj  1  1    12  1    lb.    at    $.00.      Adding 

~'  ~  these  proportional  quanti- 

^     '^  ties,    we   find   that    they 

would  form  a  mixture  of  6  pounds.     And  since  the  required  mixture 

Is  y  ^  12  times  6  pounds,  we  multiply  each  of  the  proportional  terms 

by  12,  and  obtain  for  the  required  quantities,  36  lb.  at  $.30,  24  lb.  at 

$.45,  and  1 2  lb.  at  $.60.     Hence  the  following 

Rule.  Fi7id  the  proportional  numbers  as  in  Case  IT  or  Case 
III,  Divide  the  given  quantify  hy  the  sum  of  the  proportional 
quantities,  and  multiply  each  of  the  proportional  quantities  hy  the 
quotient  thus  obtained. 


EXAMPLES   FOR   PRACTICE. 

1.  A  grocer  has  coffee  worth  8  cts.,  16  cts.,  and  24  cts.  per 
pound  respectively ;  how  much  of  each  kind  must  he  use,  to  fill  a 
cask  holding  240  lb,  that  shall  be  worth  20  cts.  a  pound  ? 

Ans.  40  lb.  at  8  cts.,  40  lb.  at  16  cts.,  and  160  lb.  at  24  cts. 

2.  A  man  bought  calves,  sheep,  and  lambs,  154  in  all,  for  $154. 
He  paid  $3 J  for  each  calf,  $li  for  each  sheep,  and  $J  for  each 
lamb ;  how  many  did  he  buy  of  each  kind  ? 

Ans.  14  calves,  42  sheep,  and  98  lambs. 

3.  A  man  paid  $165  to  55  laborers,  consisting  of  men,  women, 
and  boys ;  to  the  men  he  paid  $5  a  week,  to  the  women  $1  a  week, 
and  to  the  boys  $}  a  week ;  how  many  were  there  of  each  ? 

Ans,  30  men,  5  women,  and  20  boys. 


INVOLUTION-  379 


INVOLUTION. 

OSf^*  A  Power  is  the  product  arising  from  multiplying  a 
number  by  itself,  or  repeating  it  any  number  of  times  as  a  factor. 
^  G4L0^  Involution  is  the  process  of  raising  a  number  to  a  given 
power. 

649.   The  Square  of  a  number  is  its  second  power. 

643.  The  Cube  of  a  number  is  its  third  power. 
643*    In  the  process  of  involution,  we  observe, 

I.  That  the  exponent  of  any  power  is  equal  to  the  number  of 
times  the  root  has  been  taken  as  a  factor  in  continued  multiplica- 
tion.    Hence 

II.  The  product  of  any  two  or  mora  powers  of  the  same  num- 
ber is  the  power  denoted  by  the  sum  of  their  exponents,  and 

III.  If  any  power  of  a  number  be  raised  to  any  given  power, 
the  result  will  be  that  power  of  the  number  denoted  by  the  pro- 
duct of  the  exponents. 

1.  What  is  the  5th  power  of  6  ? 

Analysis.  We 
multiply  6  by  it- 
self, and  this  pro- 
duct by  6,  and  so 
on,  until  6  has 
been  taken  5  times 
in  continued  mul- 
tiplication;  the  final  product,  7776,  is  the  power  required,  (I).  Or, 
we  may  first  form  the  2d  and  3d  powers  •  then  the  product  of  these 
two  powers  will  be  the  5th  power  required,  (11). 

2.  What  is  the  6th  power  of  12  ? 

Analysis.     Wg  find  the  cube  of 
^-^   =  i^"^^  the  second  power,  which  must  be 

144»  =  2985984,  Ans.       .^e  6th  power,  (III). 

644.  Hence  for  the  involution  of  numbers  we  have  the  fol* 
lowing 


OPERATION. 

x6 

X 

6 

X  6    X 
Or 

6  = 

7776,  Ana. 

6 

X 

6 

=  6^  = 

36 

86 

X 

6 

=  6'  = 

216 

X  6' 

=:: 

6^ 

=  216  X  36 

=  7776,  Ans. 

880  INVOLUTION. 

KuLE,  I.  Multiply  the  given  number  hi/  itself  in  continued 
mulHplicution,  till  it  has  been  taken  as  tnany  times  as  a  factor  as 
there  are  v.vits  in  the  exponent  of  the  required  power.      Or, 

IL  Multipl}/  together  two  or  more  powers  of  the  given  number, 
the  sum  of  whose  exponents  is  equal  to  the  exponent  of  the  required 
power.     Or, 

III    Raise  some  power  of  the  given  number  to  such  a  power 

that  the  product  of  the  two  exponents  shall  be  equal  to  the  exponent 

of  the  required  power, 

NoTKS.  —  1.  A  fraction  is  involved  to  any  power  by  involving  each  of  its 
terms  separately  to  the  required  power. 

2.  Mixed  numbers  should  be  reduced  to  improper  fractions  before  involution. 

3.  When  the  number  to  be  involved  is  a  decimal,  contracted  multiplication 
may  be  applied  with  great  advantage. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  square  of  79  ?  Ans.  6241. 

2.  What  is  the  cube  of  25.4?  Ans.  16387.064. 

3.  What  is  the  square  of  1450  ? 

4.  Eaise  16|  to  the  4th  power.  Ans.  79659|f  I.' 

5.  Eaise  2  to  the  20th  power.  Ans.  1048576, 

6.  Kaise  .4378565  to  the  8th  power,  reserving  5  decimals.  \ 

Ans.  .00135 -t: 

7.  Raise  1.052578  to  the  6th  power,  reserving  4  decimals. 

Ans.  1.3600  db. 

8.  Involve  .029  to  the  5th  power  ? 

Ans.  .000000020511149. 
Find  the  value  of  each  of  the  following  expressions : 

9.  4.367*.  Ans.  363.691178934721. 

10       (1)3. 

11.  (;2|/. 

12    4.G»  X  25' 
13.  (6|y  — 7.25*. 

15.  I  of  (1)3  of  (^^y, 

Note.—  Cancel  like  powers  of  the  same  factor. 

16.  7«-f-3.08. 
17    (4^  X  5«  X  12«)  -^  (4^  X  10*  X  32). 


Ans.   If  3. 

Ans.  mfil. 

Ans.  1520875. 

14.  (8J/  X  2.5^ 

Ans.  5|. 

Am,   1200 

EVOLUTION.  381 


E^^OLUTION. 

64:«>.  A  Root  is  a  factor  repeated  to  produce  a  power;  thus, 
in  the  expression  7x7x7  =  34.3,  7  is  the  root  from  which  the 
power,  343,  is  produced. 

64®.  Evolution  is  the  process  of  extracting  the  root  of  a 
number  considered  as  a  power;  it  is  the  reverse  of  Involution. 

Any  number  whatever  may  be  considered  a  power  whose  root 
is  to  be  extracted. 

04T.   A  Rational  Root  is  a  root  that  can  be  exactly  obtained. 

648.  A  Surd  is  an  indicated  root  that  can  not  be  exactly  ob- 
tained. 

649.  The  Radical  Sign  is  the  character,  ^,  which,  placed 
before  a  number,  indicates  that  its  root  is  to  be  extracted. 

6«S0.  The  Index  of  the  root  is  the  figure  placed  above  the 
radical  sign,  to  denote  what  root  is  to  be  taken.  When  no  index 
is  written,  the  index,  2,  is  always  understood. 

6«dl*  The  names  of  roots  are  derived  from  the  corresponding 
powers,  and  are  denoted  by  the  indices  of  the  radical  sign.  Thus, 
•s/lOO  denotes  the  square  root  of  100;  \^1U0  denotes  the  cube 
roo^of  100;   v^  1 00  denotes  the /oi^r^/i   root  of    100;  etc. 

6^S«  Evolution  is  sometimes  denoted  by  a  fractional  exponent, 
the  name  of  the  root  to  be  extracted  being  indicated  by  the  deno- 
minator. Thus,  the  square  root  of  10  may  be  written  10  ;  the 
cube  root  of  10,  10  ,  etc. 

6«S3«  Fractional  exponents  are  also  used  to  denote  both  invo- 
lution and  evolution  in  the  same  expression,  the  numerator  indi- 
cating the  power  to  which  the  given  number  is  to  be  raised,  and 
the  denominator  the  root  of  the  power  which  is  to  be  taken ;  thus, 

7    denotes  the  cube  root  of  the  second  power  of  7,  and  is  the 

same  as  >/V)  so  also  7^  =  \/7^ 

6«>4.  In  extracting  any  root  of  a  number,  any  figure  or  figures 
may  be  regarded  as  tens  of  the  next  inferior  order.  Thus,  in 
2546,  the  2  may  be  considered  as  tens  of  the  3d  order,  the  25  as 
tens  of  the  second  oyder,  or  the  254  as  tens  of  the  first  order. 


882  EVOLUTION. 

SQUARE   ROOT. 

653.  The  Square  Root  of  a  number  is  one  of  the  two  equal 
factors  that  produce  the  number.  Thus,  the  square  root  of  64  is 
8,  for  8  X  8  =  64. 

To  derive  the  method  of  extracting  the  square  root  of  a  num- 
ber, it  is  necessary  to  determine 

1st.  The  relative  number  of  places  in  a  number  and  its  square  root. 

2d.  The  relations  of  the  figures  of  the  root  to  the  periods  of 
the  number. 

3d.  The  law  by  which  the  parts  of  a  number  are  combined  in 
the  formation  of  its  square ;  and 

4th.  The  factors  of  the  combinations. 

6«56«  The  relative  number  of  places  in  a  given  number  and 
its  square  root  is  shrwn  in  the  following  illustrations. 

Roots.  Squares. 

i  1 

9  81 

99  98,01 

999  99,80,01 

From  these  examples  we  perceive 

1st.  That  a  root  consisting  of  1  place  may  have  1  or  2  places  in  the 
square. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root  adds  2 
places  to  the  square.     Hence, 

I.  If  we  point  off  a  numher  into  two-figure  'periods^  commencwg 
at  the  right  hand,  the  number  of  fall  periods  avd  the  left  hand 
full  or  partial  period  loill  indicate  the  numher  of  places  in  the 
square  roof. 

To  ascertain  the  relations  of  the  several  figures  of  the  root  to  the 
periods  of  the  number,  observe  that  if  any  number,  as  2345,  be  de- 
composed at  pleasure,  the  squares  of  the  left  hand  parts  will  be  ro 
lated  in  local  value  as  follows : 

20002  ^   4  00  00  00 

23002  =   5  29  00  00 

23402  ==  ^  47  56  00 

23452  ^   5  49  90  25 :  Hence, 

II.  The  square  of  the  first  figure  of  the  root  is  contained  xchoUy 
in  the  first  period  of  the  power  ;  the  square  of  the  first  two  figures 


Roots. 

Squares. 

1 

1 

10 

1,00 

100 

1,00.00 

1000 

1,00,00,00 

I 


SQUARE  ROOT.  383 


of  the  root  is  contained  wholly  in  the  first  two  periods  of  the  power  ; 
and  so  on. 

Note. —The  periods  and  figures  of  the  root  are  counted  from  the  left  hand. 

The  combinations  in  the  formation  of  a  square  may  be  shown  as 
follows : 

If  we  take  any  number  consisting  of  two  figures,  as  43,  and  decom- 
pose it  into  two  parts,  40  +  3,  then  the  square  of  the  number  may 
be  formed  by  multiplying  both  parts  by  each  part  separately :  thus, 

40  +  3 
40  4-  3 

120  +  9 
1600  +  120 

43«  =  1600  +  240  +  9  =  1849. 
Of  these  combinations,  we  observe  that  the  first,  1600,  is  the  square 
of  40  ,  the  second,  240,  is  twice  40  multiplied  by  3  ;  and  the  third,  9, 
is  the  square  of  3.     Hence, 

III.  The  square  of  a  number  composed  of  tens  and  units  is 
equal  to  the  square  of  the  tens,  plus  twice  the  tens  multiplied  hy  the 
units  J  plus  the  square  of  the  units. 

By  observing  the  manner  in  which  the  square  is  formed,  we  per- 
ceive that  the  unit  figure  must  always  be  contained  as  a  factor  in 
both  the  second  and  third  parts ;  these  parts  taken  together,  may 
therefore  be  factored,  thus,     240  +  9  ==  (80  +  3)    X   3.     Hence, 

lY.  If  the  square  of  the  tens  he  subtracted  from  the  entire 
square,  the  remainder  will  he  equal  to  tiolce  the  tens  plus  the  units 
multiplied  hy  the  units. 

1.  What  is  the  square  root  of  5405778576  ? 

OPERA  noN.  Analysis.     Pointing  ofi"  the 

5405778576  (  73524         gi^^n  number  into  periods  of 
49  two  figures  each,  the  5  periods 

show  that  there  will  be  5  fig- 
ures in  the  root,  (I).  Since 
the  square  of .  the  first  figure 
of  the  root  is  always  contained 
wholly  in  the  first  period  of 
the  power,  (II),  we  seek  for  the 

iT^^yTTi  cooi-t^  greatest  square  in  the  first  pe" 

14/044  588176  ^.    .     m       w  i,  fi  a  \. 

588176  nod,    54,   which  we   find   by 

trial  to  be  49,  and  we  place 


143 

505 
429 

1465 

7677 
7325 

14702 

35285 
29404 

884  EVOLUTION. 

its  root,  7,  as  the  first  figure  of  the  required  root,  and  regard  it  aS 
tens  of  the  next  inferior  order,  (II).  We  now  subtract  49,  the 
square  of  the  first  figure  of  the  root,  from  the  first  period,  54,  and 
bringing  down  the  next  period,  obtain  505  for  a  remainder.  And 
since  the  square  of  the  first  two  figures  of  the  root  is  contained  wholly 
in  the  first  two  periods  of  the  power,  (II),  the  remainder,  505,  must 
contain  at  least  twice  the  first  figure  (tens)  j)^us  the  second  fiigt. re 
(units),  multiplied  hy  the  second  figure,  (IV).  Now  if  we  could  divide 
this  remainder  by  tioice  the  first  figure  plus  the  second,  which  is  one 
of  the  factors,  the  quotient  would  be  the  second  figure,  or  the  other 
factor.  But  since  we  have  not  yet  obtained  the  second  figure,  the 
complete  divisor  can  not  now  be  employed ;  and  w^e  therefore  write 
twice  the  first  figure,  or  14,  at  the  left  of  505  for  a  tried  divisor,  re- 
garding it  as  tens.  Dividing  the  dividend,  exclusive  of  the  right 
hand  figure,  by  14,  we  obtain  3  for  the  second,  or  trial  figure  of  the 
root,  which  we  annex  to  the  trial  divisor,  14,  making  143,  the  com- 
plete divisor.  Multiplying  the  complete  divisor  by  the  trial  figure 
3,  and  subtracting  the  product  from  the  dividend,  we  have  '^6  for  a 
remainder.  We  have  now  taken  the  square  of  the  first  two  figures  of 
the  root  from  the  first  two  periods ;  and  since  the  square  of  the  first 
three  figures  of  the  root  is  contained  wholly  in  the  first  three  periods, 
(II)  we  bring  down  the  third  period,  77.  to  the  remainder,  7G,  and 
obtain  for  a  new  dividend  7677,  which  must  contain  at  least  ticice  ihe 
two  figures  already  found  plus  the  third,  mxdtiplied  hy  the  third,  (lY). 
Therefore  to  obtain  the  third  figure,  we  must  take  for  a  new  trial 
divisor  twice  the  two  figures,  73,  considered  as  tens  of  the  next  infe- 
rior order,  which  we  obtain  in  the  operation  by  doubling  the  last  fig- 
ure of  the  last  complete  divisor,  143,  making  146.  Dividing,  we  ob- 
tain 5  for  the  next  figure  of  the  root ;  then  regarding  735  as  tens  of 
the  next  inferior  order,  we  proceed  as  in  the  former  steps,  and  thus 
continue  till  the  entire  root,  73524,  is  obtained. 

6«5T.  From  these  principles  and  illustrations  we  derive  the 
following 

Rule.  1.  Point  off  the  given  numher  into  periods  of  two  figures 
each,  counting  from  units  place  toicard  the  left  and  right, 

II.  Find  ihe  greatest  square  numher  in  the  left  hand p)criod,  and 
write  its  root  for  the  first  figure  in  the  root ;  subtract  the  square 
numher  from  the  left  hand  period,  and  to  the  remainder  hring 
dawn  the  next  period  for  a  dividend. 


SQUARE  ROOT.  §§5 

III.  At  the  left  of  the  dividend  write  twice  the  fir^t  figure  of  the 
root,  for  a  trial  divisor  ;  divide  the  dividend,  exclusive  of  its  right 
hand  figure,  hy  the  trial  divisor,  and  write  the  quotient  for  a  trial 
figure  in  the  root, 

Y\ .  Annex  the  trial  figure  of  the  root  to  the  trial  divisor  for  a 
complete  divisor ;  midtipli/  the  complete  divisor  hy  the  trial  figure 
in  the  root,  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

V.  Multiply  the  last  figure  of  the  last  complete  divisor  by  2  and 

add  the  product  ^o  10  times  the  previous  divisor,  for  a  new  trial 

divisor,  with  which  proceed  as  before. 

Notes. — 1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish 
the  trial  figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  a  cipher  to  the  trial  divisor,  and  another 
period  to  the  dividend,  and  proceed  as  before. 

3.  If  there  is  a  remainder  after  all  the  periods  have  been  brought  down, 
annex  periods  oi  ciphers,  and  continue  the  root  to  as  many  decimal  places  as 
are  required. 

4.  The  decimal  points  in  the  work  may  be  omitted,  care  being  taken  to  point 
off  in  the  root  according  to- the  number  of  decimal  periods  used. 

5.  The  square  root  of  a  common  fraction  may  be  obtained  by  extracting  the 
souare  roots  of  the  numerator  and  denominator  separately,  provided  the  terms 
are  perfect  squares;  otherwise,  the  fraction  may  first  be  reduced  to  a  decimal. 

6.  Mixed  numbers  may  be  reduced  to  the  decimal  form  before  extracting  the 
root ;  or,  if  the  denominator  of  the  fraction  is  a  perfect  square,  to  an  improper 
fraction. 

7.  The  popil  will  acquire  greater  facility,  and  secure  greater  accuracy,  by 
keeping  units  of  like  order  under  each  other,  and  each  divisor  opposite  the 
correspoading  dividend,  as  shown  in  the  operation. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  square  root  of  315844  ?  Ans.  562. 

2.  What  is  the  square  root  of  152399025  ?       Ans.  12345. 

3.  What  is  the  square  root  of  56280004  ?     Of  597  ?  ' 

4.  What  is  the  square  root  of  10795.21  ?  Ans.  103.9. 

5.  What  is  the  square  root  of  58.14061  ?  Ans.  7.62i. 
Find  the  values  of  the  following  expressions : 

6.  v/. 01)00316969.  A7is.  .00563. 

7.  v/3858.07694409"64.  Ans.  62.11342. 

8.  n/|.  Ans.  .745355+. 


9.  \/9^225  —  63504.  10.  \/. 126736— \/.045369. 

11.  ^\n  X  ^im^  Ans.  {3. 

12.  v/8P~x  625^  x~2^  Ans.  202500. 

33  z 


886 


EVOLUTION. 


48 

400 
384 

562 

1600 
1124 

5648 

47600 
45184 

5656 

2416* 

2262 

566 

154 
113 

CONTRACTED    METHOD. 
6*18.    1.  Find  the   square   root    of  8;  correct  to  6  decimal 
places. 

OPERATION.  Analysis.     Extracting  the  square 

12.8284274-,  Ans,      root  in  the  usual  way  until  we  have 
8  000006  obtained   the   4  places,  2.828,    the 

4  corresponding  remainder  is  2416,  and 

the  next  trial  divisor,  with  the  cipher 
omitted,  is  5656.  We  now  omit  to 
bring  down  a  period  of  ciphers  to 
the  remainder,  thus  contracting  the 
dividend  2  places ;  and  we  contract 
the  divisor  an  equal  number  of  places 
by  omitting  to  annex  the  trial  figure 
of  the  root,  and  regarding  the  right 
hand  figure,  6,  as  a  rejected  or  re- 
dundant figure.  We  now  divide  as 
__  ,--  in  contracted  division  of  decimals, 

^Q  (226),  bringing  down  each  divisor  in 

its  place,  with  one  redundant  figure 
increased  by  1  when  the  rejected  figure  is  5  or  more,  and  carrying  the 
tens  from  the  redundant  figure  in  multiplication.  We  observe  that 
the  entire  root,  2.828427+,  contains  as  manT/  places  as  there  are  places 
in  the  periods  used.     Hence  the  following 

KuLE.  I.  If  necessary^  annex  periods  of  ciphers  to  the  given 
number  J  and  assume  as  many  figures  as  tJiere  are  places  requirecl 
in  the  root ;  then  proceed  in  the  usual  manner  until  all  the  assumed 
figures  have  been  employ ed,  omitting  the  remaining  figures^  if  any, 
II.  Form  the  next  trial  divisor  as  usualy  but  omit  to  annex  to  it 
the  trial  figure  of  the  root  j  reject  one  figure  from  the  right  to  form 
each  subsequent  divisor ^  and  in  multiplying  regard  the  right  hand 
fi.gvre  of  each  contracted  divisor  as  redundant 

NoTKR. — 1.  If  the  rejected  figure  is  5  or  more,  increase  the  next  left  hand 
fj^'uie  by  1. 

2.  Alwii3''s  take  full  periods,  both  of  decimals  and  integers. 

EXAMPLES    FOR    PRACTICE. 

1.  Find  the  square  root  of  82  correct  to  the  seventh  decimal 
place.  Ans.  5.6568542  +  . 


CUBE  ROOT.  387 

2.  Find  the  square  root  of  12  correct  to  the  Beventh  decimal 
place.  Ans.  3.4G41016+. 

8.  Find  the  square  root  of  3286.9835  correct  to  the  fourth 
decimal  place.  Ans.  57.3322 -f-. 

4.  Find  the  square  root  of  .5  correct  to  the  sixth  decimal 
place.  Ans.  .745355+. 

5.  Find  the  square  root  of  6^  correct  to  the  sixth  decimal 
place.  Ans.  2.563479  +  . 

6.  Find  the  square  root  of  1.06^  correct  to  the  sixth  decimal 
place.  Ans,   1.156817+ . 

3 

7.  Find  the  value  of  1.0125^  correct  to  the  fourth  decimal 
place.  Ans.   1.0188+. 

8.  Find  the  value  of  1.023375^  correct  to  the  sixth  decimal 
place.  Ans,  1.011620  + . 

CUBE  ROOT. 

6«S9«  The  Cube  Root  of  a  number  is  one  of  the  three  equal 
factors  that  produce  the  number.  Thus,  the  cube  root  of  343  is 
7,  since  7x7x7  =  343. 

To  derive  the  method  of  extracting  the  cube  root  of  a  number, 
it  is  necessary  to  determine 

1st.  The  relative  number  of  places  in  a  given  number  and  its 
cube  root. 

2d.  The  relations  of  the  figures  of  the  root  to  the  periods  of 
the  number. 

3d.  The  law  by  which  the  parts  of  a  number  are  combined  in 
the  formation  of  a  cube ;  and 

4th.  The  factors  of  these  combinations. 

600*  The  relative  number  of  places  in  a  given  number  and 
its  cube,  is  shown  in  the  following  illustrations : 


Roots. 

Cubes. 

1 

9 

99 
999 

1 

729 

907,299 

997,002,999 

Roots. 

Cubes. 

1 

10 

100 

1000 

1 

1,000 

1,000,000 

1,000,000,000 

From  these  examples,  we  perceive, 


3g8  EVOLUTION. 

1st.  That  a  root  consisting  of  1  place  may  have  from  1  to  3  places 
in  the  cube. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root  adds  3 
places  to  the  cube.     Hence, 

I.  If  we  point  off  a  number  into  three-figure  periods,  com- 
mencing at  the  right  hand,  the  number  of  full  periods  and  the  left 
hand  full  or  partial  period  will  indicate  the  number  of  places  in 
the  cube  root. 

To  ascertain  the  relations  of  the  several  figures  of  the  root  to  the 
periods  oi  the  number,  observe  that  if  any  number,  as  5423,  be  de- 
composed, the  cubes  of  the  parts  will  be  related  in  local  value,  as 
follows : 

6000»  =  125  000  000  000 
6400»  =  157  4G4  000  000 
5420*  =159  220  088  000 
5423»  =  159  484  621  967.     Hence, 
II:    The  cube  of  the  first  figure  of  the  root  is  contained  wholly  in 
the  first  period  of  the  power  ;  the  cube  of  the  first  two  figures  of 
the  root  is  contained  wholly  in  the  first  two  periods  of  the  power; 
and  so  on 

To  learn  the  combinations  of  tens  and  units  in  the  formation  of  a 
cube,  take  any  number  consisting  of  two  figures,  as  54,  and  decom- 
pose it  into  two  parts,  50+4 ;  then  having  formed  the  square  by  656, 
III,  multiply  each  part  of  this  square  by  the  units  and  tens  of  54 
separately,  thus, 

542  ==                            502  -f-  2  X  50  X  4  +  42 
50  +  4 

502  X  4  +  2  X  50  X  42  +  43 
508+2x50^x4+  50X4^ 

543=  503+3  X  50^x4+3  X  50  X  42  +  43  =  156924 
Of  these  combinations,  the  first  is  the  cube  of  50,  the  second  is  3 
times  the  square  of  50  multiplied  by  4,  the  third  is  3  times  50  multi- 
plied by  the  square  of  4,  and  the  fourth  is  the  cube  of  4.     Hence, 

III.  The  cube  of  a  number  composed  of  ten^  and  units  is  equal 
to  the  cube  of  the  tern,  plus  three  times  the  square  of  the  tens  multi- 
plied by  the  units,  plus  three  times  the  tens  multiplied  by  the  square 
of  the  units,  plus  the  cube  of  the  units. 

By  observing  the  manner  in  which  the  cube  is  formed,  we  perceive 
that  each  of  the  last  three  parts  contains  the  units  as  a  factor ;  these 


CUBE  EOOT.  38& 

parts,  considered  as  one  number,  may  therefore  be  separated  into  two 
factors,  thus, 

(3  X  502  +  3  X  50  X  4  +  42)  X  4     Hence, 

TV.  If  the  cube  of  the  tens  he  subtracted  from  the  entire  cube^ 
the  remainder  will  be  composed  of  two  factors^  one  of  which  will  be 
three  times  the  square  of  the  tens  plus  three  times  the  tens  multipUed 
by  the  units  plus  the  square  of  the  units  ;  and  the  other j  the  units, 

1.  What  is  the  cube  root  of  145780726447  ? 

OPERATION. 

145780726447  ( 5263,  Ans. 
I  II  125 


152  304 


1566        9396 


7500        20780 
7804        15608 


811200      5172726 
820596      4923576 


83002800    249150447 
15783    47349      83050149    249150447 

Analysis.  Pointing  off  the  given  number  into  periods  of  3  figures 
each,  the  four  periods  show  that  there  will  be  four  figures  in  the  root, 
(I).  Since  the  cube  of  the  first  figure  of  the  root  is  contained  wholly 
in  the  first  period  of  the  power,  (II),  we  seek  the  greatest  cube  in  the 
first  period,  145,  which  we  find  by  trial  to  be  125,  and  we  place  its 
root,  5,  for  the  first  figure  of  the  required  root,  and  regard  it  as  tens 
of  the  next  inferior  order,  (654).  We  now  subtract  125,  the  cube 
of  this  figure,  from  the  first  period,  145,  and  bringing  down  the  next 
period,  obtain  20780  for  a  dividend.  And  since  the  cube  of  the  first 
two  figures  of  the  root  is  contained  wholly  in  the  first  two  periods 
of  the  powor,  (II),  the  dividend,  20780,  must  contain  at  least  the 
product  of  the  two  factors,  one  of  which  is  three  times  the  square  of 
the  first  figure  (tens),  plus  three  times  the  first  figure  multiplied  by 
the  second  (units),  ^Zz/5  the  square  of  the  second  ;  and  the  other,  tho 
second  figure  (IV).  Now  if  we  could  divide  this  dividend  by  the  first 
of  these  factors,  the  quotient  would  be  the  other  fuctor,  or  the  second 
figure  of  the  root.  But  as  the  first  factor  is  composed  in  part  of  the 
second  figure,  which  we  have  not  yet  found,  we  can  not  now  obtain  the 
complete  divisor ;  and  we  therefore  write  three  times  tho  square  of 
the  first  figure,  regarded  as  tens,  or  50^  X  3  =  7500,  at  the  left  of  the 
dividend,  for  a  trial  divisor.  Dividing  the  dividend  by  the  trial 
divisor,  we  obtain  2  for  the  second,  or  trial  figure  of  the  root.  To 
33* 


390  EVOLUTION. 

complete  the  divisor,  we  must  add  to  the  trial  divisor,  as  a  correction, 
three  times  the  tens  of  the  root  already  found  multiplied  by  the  units, 
plus  the  square  of  the  units,  (lY).  But  as  50  X  3  X  2  -f  2^  r=  (50  X 
3  -f-  2)  X  2,  we  annex  the  second  figure,  2,  to  three  times  the  first 
figure,  5,  and  thus  obtain  50  X  3  -f  2  =  152,  the  first  factor  of  the 
correction,  which  we  write  in  the  column  marked  I.  Multiplying 
this  result  by  the  2,  we  have  304,  the  correction,  which  we  write  in 
the  column  marked  II.  Adding  the  correction  to  the  trial  divisor,  we 
obtain  7804,  the  complete  divisor.  Multiplying  the  complete  divisor 
by  the  trial  figure  of  the  root,  subtracting  the  product  from  the 
dividend,  and  bringing  down  the  next  period,  we  have  5172726  for 
a  dividend. 

We  have  now  taken  the  cube  of  the  first  two  figures  of  the  root 
considered  as  tens  of  the  next  inferior  order,  from  the  first  three 
periods  of  the  number ;  and  since  the  cube  of  the  first  three  figures 
of  the  root  is  contained  wholly  in  the  first  three  periods  of  the  power, 
(II),  the  dividend,  5172726  must  contain  at  least  the  product  of  the 
two  factors,  one  of  which  is  tliret  times  the  square  of  the  first  two 
figures  of  the  root  (regarded  as  tens  of  the  next  order)  plus  three 
times  the  first  two  figures  multiplied  hy  the  third,  plus  the  square  of  the 
third;  and  the  other,  the  third  figure,  (IV).  Therefore,  to  obtain  the 
third  figure,  we  must  use  for  a  trial  divisor  three  times  the  square  of 
the  first  two  figures,  52,  considered  as  tens.  And  we  observe  that  the 
significant  part  of  this  new  trial  divisor  may  be  obtained  by  adding 
the  last  complete  divisor,  the  last  correction,  and  the  square  of  the 
last  figure  of  the  root,  thus : 

7804  =  (502  X  3)  +    (50  X  3  X  2)  +  22 

304  =  50  X  3  X  2  +  22 

4= 22 

8n2^  (502  +  lOO'x  2  +22) X  3  =  522  X  3 

This  number  is  obtained  in  the  operation  without  re-writing  the 
parts,  by  adding  the  square  of  the  second  root  figure  mentally,  and 
combining  units  of  like  order,  thus :  4,  4,  and  4  are  12,  and  we  write 
the  unit  figure,  2,  in  the  new  trial  divisor ;  then  1  to  carry  and  0 
is  1 ;  then  3  and  8  are  11,  etc.  Annexing  two  ciphers  to  the  8112, 
because  52  is  regarded  as  tens  of  the  next  order,  and  dividing  by  this 
new  trial  divisor,  811200,  we  obtain  6,  the  third  figure  in  the  root. 
To  complete  the  second  trial  divisor,  after  the  manner  of  completing 
the  first,  we  should  annex  the  third  figure  of  the  root,  6,  to  three 
times  the  former  figures,  52,  for  the  first  factor  of  the   correction. 


CUBE  ROOT.  «  391 

But  as  we  have  in  column  I  three  times  5  with  the  2  annexed,  or  152, 
we  need  only  multiply  the  last  figure,  2,  by  3,  and  annex  the  third 
figure  of  the  root,  6,  which  gives  1566,  the  first  factor  of  the  correc- 
tion sought,  or  the  second  term  in  column  I.  Multiplying  this  number 
by  the  6,  we  obtain  9396,  the  correction  sought ;  adding  the  correction 
to  the  trial  divisor,  we  have  820596,  the  complete  divisor ;  multiplying 
the  complete  divisor  by  the  6,  subtracting  the  product  from  the  divi- 
dend, and  bringing  down  the  next  period,  we  have  249150447  for  a 
new  dividend  We  may  now  regard  the  first  three  figures  of  the  root, 
526,  as  tens  of  the  next  inferior  order,  and  proceed  as  before  till  the 
entire  root,  5263,  is  extracted. 

6G1*  From  these  principles  and  illustrations  we  deduce  the 
following 

Rule.  I.  Point  off  the  given  number  info  periods  of  three 
figures  each,  counting  from  units" place  toward  the  left  and  right. 

II.  Find  the  greatest  cube  that  does  not  exceed  the  left  hand 
period  J  and  icrite  its  root  for  the  first  figure  in  the  required  root; 
subtract  the  cube  from  the  left  hand  period^  and  to  the  remainder 
bring  down  the  next  period  for  a  dividend. 

III.  At  the  left  of  the  dividend  write  three  times  the  square  of 
the  first  figure  of  the  root,  and  annex  two  ciphers,  for  a  trial  di- 
visor;  divide  the  dividend  by  the  trial  divisor,  and  write  the  quo- 
tient for  a  trial  figure  in  the  root. 

lY.  Annex  the  trial  figure  to  three  times  the  former  figure,  and 
write  the  result  in  a  column  marhed  I,  one  line  below  the  trial 
divisor ,  multiply  this  term,  by  the  trial  figure,  and  write  the 
product  on  the  same  line  in  a  column  marked  II;  add  this  term 
as  a  correction  to  the  trial  divisor,  and  the  result  will  be  the  com- 
plete divisor. 

V.  Multiply  the  complete  divisor  by  the  trial  figure ;  sidttract 
the  product  from  the  dividend,  and  to  the  remainder  bring  dozen 
the  next  period  for  a  new  dividend. 

YI.  Add  the  square  of  the  last  figure  of  the  root,  the  last  term 
in  column  II,  and  the  complete  divisor  together,  and  annex  two 
ciphers^  for  a  new  trial  divisor  j  with  which  obtain  another  trial 
figure  in  the  root. 


392  EVOLUTION. 

YII.  Multiply  the  unit  figure  of  the  last  term  in  column  I  hy 
3,  and  annex  the  trial  figure  of  the  root  for  the  next  term  of 
column  I;  multiply  this  result  hy  the  trial  figure  of  the  root  for 
the  next  term  of  column  II ;  add  this  term  do  the  trial  divisor  for 
a  complete  divisor,  icith  which  proceed  as  hefore. 

Notes. — 1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish 
the  trial  figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the  trial  divisor, 
nnd  another  period  to  the  dividend;  then  proceed  as  before  with  column  I,  an* 
nexlng  both  cipher  and  trial  figure. 

EXAMPLES   FOR   PRACTICE, 

1.  What  is  the  cube  root  of  389017  ?  Ans,  73. 

2.  What  is  the  cube  root  of  44361864  ?  Ans.  354. 

3.  What  is  the  cube  root  of  10460353203  ?        Ans.  2187. 

4.  What  is  the  cube  root  of  98867482624  ?        Ans.  4624» 

5.  What  is  the  cube  root  of  30.625  ?         Ans.  3.12866  +. 

6.  What  is  the  cube  root  of  111  J  ?  Ans    4.8076  f . 

7.  What  is  the  cube  root  of  .000148877?  A7is.  .053. 
Find  the  ^'alues  of  the  following  expressions. 

8.  ^12'2615327232y  Ans.  4968. 

9.  ^n7i34^W^  Ans.  8. 
10.  Va¥30¥'?                                                       Ans.  1156. 

•      ^^I)D5    ^    ^   3TT9   *  ^'^^'     B5' 

12.  How  much  does  the  sum  of  the  cube  roots  of  50  and  31 
exceed  the  cube  root  of  their  sum?  Ans,  2.4986  +. 

CONTRACTED    METHOD. 

G63.  In  applying  contracted  decimal  division  to  the  e  -tac- 
tion of  the  cube  root  of  numbers,  we  observe, 

1st.  For  each  new  figure  in  the  root,  the  terms  in  the  operati'  a 
extend  to  the  right  3  places  in  the  column  of  dividends,  2  placv.s 
in  the  column  of  divisors,  and  1  place  in  column  I.     Hence, 

2d.  If  at  any  point  in  the  operation  we  omit  to  bring  down  new 
periods  in  the  dividend,  we  must  shorten  each  succeeding  divisor 
1  place,  and  each  succeeding  term  in  column  I,  2  places. 

1.  What  is  the  cube  root  of  189,  correct  to  8  decimal  places  ? 


CUBE  ROOT. 


393 


II 


OPERATION. 

|5.73879355dz,  Ans. 

189.000000 
125 


Analysis.  We 
proceed  by  the  usual 
method  to  extract 
the  cube  root  of  the 
given  number  until 
we  have  obtained 
the  three  figures, 
5.73 :  the  corres- 
ponding remainder 
is  867483,  and  the 
next  trial  divisor 
with  the  ciphers 
omitted  is  984987. 
We  now  omit  to 
bring  down  a  period 
of  ciphers,  thus  con- 
tracting the  divid- 
end 3  places ;  and 
we  contract  the  di- 
visor an  equal  num- 
ber of  places  by 
emitting  to  annex  the  two  ciphers,  and  regarding  the  right  hand 
figure,  7,  as  a  redundant  figure.  Then  dividing,  we  obtain  8  fcH*  the 
next  figure  of  the  root.  To  complete  the  divisor,  we  obtain  a  correc- 
tion, 1375,  contracted  2  places  by  omitting  to  annex  the  trial  figure 
of  the  root,  8,  to  the  first  factor,  1719,  and  regarding  the  right  hand 
figure,  9,  as  redundant  in  multiplying.  Adding  the  contraction  to 
the  contracted  divisor,  we  have  the  complete  divisor,  986362,  the  right 
hand  figure  being  redundant.  Multiplying  by  8  and  subtracting  the 
product  from  the  dividend,  we  have  78393  for  a  new  dividend.  Then 
to  form  the  new  trial  divisor,  we  disregard  the  square  of  the  root 
figure,  8,  because  this  square  consists  of  the  same  orders  of  units  as 
the  two  rejected  places  in  the  divisor;  and  we  simply  add  the  cor- 
rection, 1375,  and  the  complete  divisor,  986362,  and  rejecting  1  figure, 
thus  obtain  98774,  of  which  the  right  hand  figure,  4,  is  redundant. 
Dividing,  we  obtain  7  for  the  next  root  figure.  Rejecting  2  places 
from  the  last  term  in  column  I,  we  have  17  for  the  next  contracted 
term  in  this  column.  We  then  obtain,  by  the  manner  shown  in  the 
former  step,  the  correction  12,  the  complete  divisor,  98786,  the  prod- 
uct, 69150,  and  the  new  dividend,  9243.    We  then  obtain  the  new  trial 


157 

1099 
5139 

7500 
8599 

64  000 
60193 

1713 

974700 
979839 

3  807000 
2  939517 

1719 

1375 

984987 
986362 

867483* 
789090 

17 

12 

98774 
98786 

78393 
69150 

9880 

9243 

8892 

988 

351 

296 

99 

55 

50 

10 

5 
5 

894  ETOLUTION. 

divisor,  9880;  and  as  column  I  is  terminated  by  rejecting  the  two 
places,  17,  we  continue  the  contracted  division  as  in  square  root,  and 
thus  obtain  the  entire  root,  5.73879355  db,  which  is  correct  to  the  last 
decimal  place,  and  contains  as  many  places  as  there  are  places  in  the 
periods  used.     Hence  the  following 

Rule.  I.  If  necessary ,  amiex  ciphers  to  the  given  number j  and 
assume  as  many  figures  as  there  are  places  required  in  the  root ; 
then  proceed  by  the  usual  method  until  all  the  assumed  figures  have 
been  employed. 

II.  Form  the  next  trial  divisor  as  usual,  but  omit  to  annex  the 
two  ciphers^  and  reject  one  place  in  forming  each  subsequent  tinal 
divisor, 

III  In  completing  the  contracted  divisor Sy  omit  at  first  to  annex 
the  trial  figure  of  the  root  to  the  term  in  column  I,  and  reject  2 
2ilaces  in  forming  each  succeeding  term  in  this  column. 

lY.  In  multiply'mg,  regard  the  right  hand  figure  of  each  con- 
tracted term,  in  column  I  and  in  the  column  of  divisors,  as  redund- 
ant. 

Notes. — 1.  After  the  contraction  commences,  the  square  of  the  last  root  figure 
is  disrej2;arded  in  forming  the  new  trial  divisors. 
2.  Employ  oxAy  full  periods  in  the  number. 

EXAMPLES   FOR   PRACTICE. 

1.  Find  the  cube  root  of  24,  correct  to  7  decimal  places. 

Ans.  2.8844992  ±. 

2.  Find  the  cube  root  of  12000.812161,  correct  to  9  decimal 
places.  Ans.  22.894801334  db. 

8.  Find  the  cube  root  of  .171467,  correct  to  9  decimal  places. 

Ans.  .555554730  ±. 

4.  Find  the  cube  root  of  2. 42999  correct  to  5  decimal  places. 

Ans.  1.34442±. 

5.  Find  the  cube  root  of  19.44,  correct  to  4  decimal  places. 

Ans.  2.6888    ±. 

6.  Find  the  value  of  v^l"  to  6  places.  Ans.  .941035  ±. 

7.  Find  the  value  of  ^.571428  to  9  places. 

Ans.  .829826686  ±. 


ROOTS  OF  ANY  DEGREE.  395. 

8.  Find  the  value  of  VlU8G74325^  to  7  places. 

Ans,  1.057023  zfc. 
5 

9.  Eind  the  value  of  1.053  to  7  places. 

Ans.  1.084715  ±. 

ROOTS   OP   ANY   DEGREE. 
063.    Any  root  whatever  may  be  extracted  by  means  of  the 
square  and  cube  roots,  as  will  be  seen  in  the  two  cases  which  follow. 

CASE   I. 

6G4.  When  the  index  of  the  required  root  contains 
no  other  factor  than  2  or  3. 

We  have  seen  that  if  we  raise  any  power  of  a  given  number  to 
any  required  power,  the  result  will  be  that  power  of  the  given 
number  denoted  by  the  product  of  the  two  indices,  (64:3,  III). 
Conversely,  if  we  extract  successively  two  or  more  roots  of  a  given 
number,  the  result  must  be  that  root  of  the  given  number  denoted 
by  the  product  of  the  indices. 

1.  What  is  the  6th  root  of  2176782336  ? 

OPERATION.  Analysis.     The  index  of  the 

6  =  2x3  required  root  is  6  =  2x3;  we 

\/ 2176782336  =  46657  therefore  extract  the  square  root 

v^ 46656  =  36    Ans,         ^^  *he   given   number,  and  the 

cube  root  of  this  result,  and  ob- 

Or  . 

'  tain  36,  which  must  be  the  6tli 

%^2176782336  =  1296  root  required.     Or,  we  first  find 

\/l296  =  36,  Ans.         the  cube  root  of  the  given  num- 

ber, and  then  the  square  root  of 
the  result,  as  in  the  operation.     Hence  the  following 

KuLE.  Separate  the  index  of  the  required  root  iiito  its  prime 
factors,  and  extract  successively  the  roots  indicated  hy  the  several 
factors  obtained ;  the  final  result  will  he  the  required  root, 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  6th  root  of  6321363049  ?  Ans.  43. 

2.  What  is  the  4th  root  of  5636405776  ?  Am,  274. 


396  EVOLUTION. 

3.  What  is  the  8th  root  of  1099511627776  ?  Ans,  82. 

4.  What  is  the  6th  root  of  25632972850442049  ?    Ans.  543. 

5.  What  is  the  9th  root  of  1.577635  ?        Ans.  1.051963 +. 

Note. — Extract  the  cube  root  of  the  cube  root  by  the  contracted  method, 
carrying  the  root  iu  each  operation  to  6  decimal  places  only. 

6.  What  is  the  12th  root  of  16.3939  ?  Ans.  1.2624 +  . 

7.  What  is  the  18th  root  of  104.9617  ?         Am.  1.2950+. 

CASE    II. 

G&S.  When  the  index  of  the  required  root  is  prime, 
or  contains  any  other  factor  than  2  or  3. 

To  extract  any  root  of  a  number  is  to  separate  the  number  into 
as  many  equal  factors  as  there  are  units  in  the  index  of  the  re- 
quired root ;  and  it  will  be  found  that  if  by  any  means  we  can 
separate  a  number  into  factors  nearly  equal  to  each  other,  the 
average  of  these  factors,  or  their  sum  divided  the  number  of  fac- 
tors, will  be  nearly  equal  to  the  root  indicated  by  the  number  of 
factors. 

1.  What  is  the  7th  root  of  308  ? 

OPERATION.  Analysis.     We  first 

^oQg  ^  2.59-f  fi^^  ^y  ^^s®  I'  tl^e  6t^^ 

^3Qg  __  2.044-  ^^^*»  ^^^  ^^^^  *^^®  ^'^'^ 

2.59-f-  2.04  =  4.63  root  of  308 ;  and  since 

4.63  ^       2  =  2.31,  assumed  root.  the  7th  root  must  be 

2.316         =151.93  less   than   the    former 

308  ~  151.93  =  2.0272+  and   greater  than   the 

?;.'L79'-+7'-'?9ATr''l^^^'         •      ^-         latter,  we  take  the  ave- 
15.8872  -r-  7  =  2.2596,  1st  approximation.  ^  ^^     , 
^-^ rage  of  the  two,  or  one 

fo^f  1^''r7°if '%  9^^ir,,  half  of  theirsums,2.31, 

308  —  13b.b748  =  2.253452-f-  ,      ,,  .,  ,,  \ 

2.2696  X  6  +  2.253452  =  15.871052  ^^^  ^^^^  ^*  *^®  assumed 

15.871052  -^  7  =  2.267293,  2d  approx.         '^oot.     We    next    raise 

the  assumed  root,  2.31, 
to  the  6th  power,  and  divide  the  given  number,  308,  by  the  result, 
and  obtain  2.0272+  for  a  quotient ;  we  thus  separate  308  into  7  fac- 
tors, 6  of  which  are  equal  to  2.31,  and  the  other  is  2.0272.  As  these 
7  factors  are  nearly  equal  to  each  other,  the  average  of  them  all  must 
be  a  near  approximation  to  the  7th  root.  Multiplying  the  2.31  by  6, 
adding  the  2.0272  to  the  product,  and  dividing  this  result  by  7,  we 


ROOTS  OF  ANY   DEGREE.  897 

find  the  average  to  be  2.2696,  which  is  the  first  approximation  to  the 
required  root.  We  next  divide  308  by  the  6th  power  of  2.2G06,  and 
obtain  2.253452-f-  for  a  quotient ;  and  we  thus  separate  the  given 
number  into  7  factors,  6  of  which  are  each  equal  to  2.2696,  and  the 
other  is  2.253452.  Finding  the  average  of  these  factors,  as  in  the 
former  steps,  we  have  2.267293,  which  is  the  7th  root  of  the  given 
number,  correct  to  5  decimal  places.  Hence  the  following 
.  Rule.  I.  Find  hy  trial  wine  number  nearJy  equal  to  the  re- 
quired rootj  and  call  this  the  assumed  root. 

II.  Divide  the  given  number  by  that  power  of  the  assumed  root 
denoted  by  the  index  of  the  required  root  less  1 ;  to  this  quotient 
odd  as  many  times  the  assumed  root  as  there  (ire  units  in  the 
index  of  the  required  root  less  1,  and,  divide  the  amount  by  the 
index  of  the  required  root.  The  result  will  be  the  first  approxi- 
mate root  required. 

III.  Take  the  last  approximation  for  the  assumed  root,  with 
which  proceed  as  with  the  former,  and  thus  continue  till  the  re- 
quired root  is  obtained  to  a  sufficient  degree  of  exactness. 

Notes. — 1.  The  involution  and  division  in  all  cases  will  be  much  abridged  by 
decimal  contraction. 

2.  If  the  index  of  the  required  root  contains  the  factors,  2  or  3,  we  may  first 
extract  the  square  or  cube  root  as  many  times,  successively,  as  these  factors  are 
found  in  the  index,  after  which  we  must  extract  that  root  of  the  result  which  is 
denoted  by  the  remaining  factor  of  the  index.  Thus,  if  the  15th  root  were  re- 
quired, we  should  first  find  the  cube  root,  then  the  5tii  root  of  this  result. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  20ih  root  of  617  ? 

OPERATION. 

20  =  2   X    2   X   5. 

v/617  =  24.839485+. 
v^^^4.839485  =  4.983923+. 
s/IWd^rS     =     L378206  +  .  ^ws. 

2.  What  is  the     5th  root  of  120  ? 

3.  What  is  the     7th  root  of  1.95678  ? 

4.  What  is  the  10th  root  of  743044? 

5.  What  is  the  15th  root  of  15  ? 
6  What  is  the  25th  root  of  100  ? 
.7.  What  is  the  5th  root  of  5  ? 

34 


898 


SQUARE  AND  CUBE  ROOTS. 


APPLICATIONS  OF  THE  SQUARE  AND  CUBE  ROOTS. 

666.    An  Angle  is  the  opening  between  two  lines  ^ 
that  meet  each  other. 

A 


667.   A  Right  Angle  is  an  angle  formed  by  two 
lines  perpendicular  to  each  other.     Thus,  B  A  C  is  a  right  angle. 

6685  If  an  angle  is  less  than  a 
right  angle,  it  is  acute  ;  if  greater 
than  a  right  angle,  it  is  obtuse. 
Thus,  the  angle  .on  the  right  of  the  ^ 

line  C  B  is  acute,  and  the  angle  on 
the  left  of  C  B  is  obtuse.  ^ 

669.  Parallel  Lines   are  lines  hav-         a 

ing  the  same  direction,  as  A  and  B.  b 

670.  A  Triangle  is  a  figure  having  three  sides 
and  three  angles,  as  A  B  C. 

671.  A   Right- Angled  Triangle  is  a  triangle 
having  one  right  angle,  as  at  C. 

672.  The  Hypotenuse  Is  the  side  opposite  the  ^ 
right  angle,  as  A  B. 

673.  The  Base  of  a  triangle  is  the  side  on  which  it  is  sup- 
posed to  stand,  as  A  C. 

674:.    The  Altitude  of  a  triangle  is  the  perpendicular  distance 

from  the  base,  or  the  base  produced,  to  the  angle  opposite,  as  C  B. 

Note.  —  The  altitude  of  a  right-angled  triangle  is  the  side  called  the  perpen- 
dicular. 

67o.    A  Sq[ijare  is  a  figure  having  four  equal  sides  and  four 
right  angles 

676.  A  Rectangle  or  Parallelogram 

is  a  figure  having  four  right  angles,  and  its 
opposite  sides  equal. 

677.  A  Diagonal  is  a  line  drawn 
through  a  figure,  joining  two  opposite 
angles,  as  A  C. 


APPLICATIONS. 


399 


I 


G78.  A  Circle  is  a  figure  bounded  by 
one  uniform  curved  line. 

679.  The  Circumference  of  a  circle  is 
the  curved  line  bounding  it. 

080.  The  Diameter  of  a  circle  is  a 
straight  line  passing  through  the  center,  and 
terminating  in  the  circumference. 

681,    A  Semi-Circle  is  one  half  of  a  circle. 

685.  A  Prism  is  a  solid  whose  bases  or  ends 
are  any  similar,  equal,  and  parallel  plane  figures; 
and  whose  sides  are  parallelograms. 

683.  A  Parallelepiped  is  a  solid  bounded  by 
six  parallelograms,  the  opposite  ones  of  which  are 
parallel  and  equal  to  each  other.  Or,  it  is  a  prism 
whose  base  is  a  parallelogram. 

684:,  A  Cube  is  a  solid  bounded  by  six 
equal  squares.  The  cube  is  sometimes  called  a 
Right  PruTYi, 

083.  A  Sphere  or  Globe  is  a  solid 
bounded  by  a  single  curved  surface,  which  in 
every  part  is  equally  distant  from  a  point 
within  called  its  center. 

686.  The  Diameter  of  a  sphere  is  a 
straight  line  passing  through  its  center,  and 
terminating  at  its  surface. 

687.  A  Hemisphere  is  one  half  of  a  globe  or  sphere. 

688.  Similar  Figures  and  Similar  Solids  are  such  as  have 
their  like  dimensions  proportional. 


PROBLEM   I. 

689.  To  find  eitlier  side  of  a  right-angled  triangle, 
the  other  two  sides  being  given. 

Let  us  take  any  right-angled  triangle,  as  ABC,  and  form  the 
equare,  A  E  D  C,  on  the  hypotenuse.  Now  take  a  portion,  ABC,  of 
this  square,  and  move  it  as  on  a  hinge  at  A,  until  the  points  B  and  C 


400 


SQUARE  AND  CUBE  ROOTS. 


D 

1 

F     \ 

\                             ^ 

G 

^ 

are  brought  to  the  positions  of  H 
and  E,  respectively.  Take  also 
another  portion,  D  F  C,  and  move 
it  as  on  a  hinge  at  D,  until  the 
points  F  and  C  are  brought  to  the 
positions  of  G  and  E,  respectively. 
Then  the  figure  formed  by  the 
parts  thus  moved  and  the  remain- 
ing part  will  be  composed  of  two 
new   squares,    one   on    A  B,   the 

base  of  the  triangle,  and  one  on  _ 

D  F,  which  is  equal  to  the  per- 
pendicular of  the  triangle.     Hence, 

The  square  of  the  hypotenuse  of  a  right-angled  triangle  is  equal 
to  the  sum  of  the  squares  of  the  other  two  sides. 

From  this  property  we  derive  the  following 

Rule.  I.  To  find  the  hypotenuse ;  —  Add  the  squares  of  the 
two  sides  J  and  extract  the  square  root  of  the  sum. 

II.  To  find  either  of  the  shorter  sides ;  —  Subtract  the  square 
of  the  given  side  from  the  square  of  the  hypotenuse j  and  extract  the 
square  root  of  the  remainder. 


EXAMPLES   FOR   PRACTICE. 

1.  The  top  of  a  tower  standing  22  feet  from  the  shore  of  a  river, 
is  75  feet  above  the  water,  and  256  feet  in  a  straight  line  from 
the  opposite  shore  3  required  the  width  of  the  river. 

Ans.  222.76  ft. 

2.  Two  ships  set  sail  from  the  same  port,  and  one  sails  due  east 
50  leagues,  the  other  due  north  84  leagues ;  how  far  are  they 
apart  ? 

3.  A  ladder  50  ft.  long  will  reach  a  window  30  ft.  from  the 
ground  on  one  side  of  the  street,  and  without  moving  the  foot,  will 
reach  a  window  40  ft.  high  on  the  other  side  ]  what  is  the  breadth 
of  the  street? 

4.  What  is  the  distance  through  a  cubical  block,  measured  from 
one  corner  to  the  opposite  diagonal  corner,  the  side  of  the  cube 
being  6  feet?  '    Ans.  10.39  ft 


^ 


ICATIONS.  401 


PROBLEM   II. 


690.  To  find  the  side  of  a  square  equal  in  area  to  a 
''given  rectangle. 

Note. —  This  case,  arithmetically  considered,  requires  us  to  find  a  mean  pro- 
portional between  two  given  numbers. 

The  product  of  the  sides  of  the  rectangle  will  be  the  area  which 
the  square  is  to  contain ;  hence 

EuLE.  3Iultiply  the  sides  of  the  rectangle  together,  and  extract 
the  square  root  of  the  product. 

EXAMPLES   FOR   PRACTICE. 

1.  There  is  a  field  whose  length  is  208  rods,  and  whose  breadth 
is  13  rods ;  what  is  the  length  of  the  side  of  a  square  lot  contain- 
ing an  equal  area  ?  Ans.  52  rods. 

2.  If  it  cost  $312  to  inclose  a  farm  216  rods  long  and  24  rods 
wide,  how  much  less  will  it  cost  to  inclose  a  square  farm  of  equal 
area  with  the  same  kind  of  fence  ? 

3.  What  is  the  mean  proportional  between  12  and  588  ? 

Ans,   84. 

4.  A  and  B  traded  together.  A  put  in  $540  for  480  days,  and 
received  J  of  the  gain ;  and  the  number  of  dollars  which  B  put 
in  was  equal  to  the  number  of  days  it  was  employed  in  trade. 
What  was  B's  capital  ?  Ans.  $720. 

PROBLEM   III. 

691.  To  find  the  two  sides  of  a  rectangle,  the  area 
and  the  ratio  of  the  sides  being  given. 

Note. — This  case,  arithmetically  considered,  requires  us  to  find  two  numbers 
whose  product  and  ratio  are  given 

If  we  multiply  together  the  terms  of  the  given 
ratio,  the  product  will  be  the  area  of  a  rectangle 
similar  in  form  to  the  rectangle  whose  sides  are 
required.  Now  we  perceive,  by  the  accompanying 
figures,  that  multiplying  both  sides  of  any  rect- 
angle by  2,  3,  4,  eta,  multiplies  the  area  by  the 
34*  2a 


402  SQUARE  AND  CUBE  HOOTS. 


I  j  I 


squares  of  these  numbers,  or  4,  9,  16, 
etc.  If,  therefore,  we  divide  the  given 
area  by  the  rectangle  of  the  terms  pro- 
portional to  the  required  sides,  the  quo- 
tient will  be  the  square  of  that  number 
which  must  be  multiplied  into  these  pro- 
portional terms  to  produce  the  required 
sides. 

Hence  the  following 

E.ULE.  I.  Divide  the  given  area  hy  the  product  of  the  terms 
proportional  to  the  sideSy  and  extract  the  square  root  of  the  quotient. 

II.  Multiply  the  root  thus  obtained  hy  each  proportional  term  ; 
the  products  will  he  the  corresponding  sides. 

EXAMPLES   FOR   PRACTICE. 

1.  The  sides  of  a  rectangle  containing  432  square  feet  are  as  4 
to  3 ;  required  the  length  and  breadth. 

Ans.  Length,  24  feet;  breadth,  18  feet. 

2.  Separate  23  into  two  factors  which  shall  be  to  each  other  as 
2  to  3.  Ans.  3.91578  +  ;  5.87367  +• 

3.  It  is  required  to  lay  out  283  A.  2  E.  27  P.  of  land  in  the 
form  of  a  rectangle  whose  length  shall  be  3  times  the  width; 
what  will  be  the  dimensions  ? 

Note.  —  The  proportional  terms  are  3  :  1. 

Ans.  369  rods;  123  rods. 

PROBLEM   IV. 

69^.  To  find  the  radius,  diameter,  or  circumference 
of  a  circle,  the  ratio  of  its  area  to  a  known  circle  being 
given. 

All  examples  of  this  class  relating  to  circles,  may  be  solved  by 
means  of  the  following  property  :  — 

The  areas  of  two  circles  are  to  each  other  as  the  squares  of  their 

radiij  diameters^  or  circumferences. 

Note. — This  property  of  the  circle  is  only  a  particular  cnse  of  a  more  general 
principle,  viz. :  That  the  areas  of  similar  figures  are  to  each  other  as  the  squares 
of  their  like  dimensions.  This  principle  is  rigidly  demonstrated  in  Geometry, 
but  cannot  be  easily  proved  here. 


iffpLICATIONS-  403 

EXAMPLES   FOR   PRACTICE. 

1.  The  radius  of  a  circle  containing  28.2744  sq.  ft.,  is  6  ft. ; 
what  is  the  radius  of  a  circle  containing  175.7150  sq.  ft.  ? 

28.2744  :  175.7150  =  6=^  :  ()  =  225,  square  of  radius  re- 
quired.     Hence,  v^225  =  15,  Ans. 

2.  If  it  cost  $75  to  inclose  a  circular  pond  containing  a  cer- 
tain area,  how  much  will  it  cost  at  the  same  rate  to  inclose  an- 
other, containing  5  times  the  area  of  the  first?     Ans,  $167.70. 

3.  If  a  cistern  6  feet  in  diameter  hold  80  barrels  of  water,  what 
must  be  the  diameter  of  a  cistern  of  the  same  depth  to  hold  1200 
barrels  ? 

4.  If  a  pipe  1.5  in.  in  diameter  will  fill  a  cistern  in  5  h.,  what 
must  be  the  diameter  of  a  pipe  that  will  fill  the  same  cistern  in 
55  min.  6  sec.  ?  Am.  3.5  in. 

PROBLEM   V. 

693.  To  find  the  side  of  a  cube,  the  solid  contents 
being  given. 

Note. — This  case,  arithmetically  considered,  requires  us  to  separate  a  number 
into  three  equal  factors. 

The  solid  contents  of  a  cube  are  found  by  cubing  the  length  of 
one  side;  hence. 

Rule.     Extract  the  aihe  root  of  the  given  contents. 
EXAMPLES   FOR   PRACTICE. 

1.  What  must  be  the  length  of  the  side  of  a  cubical  bin  that 
shall  contain  the  same  quantity  as  one  that  is  24  ft.  long,  18  ft. 
wide,  and  4  ft.  deep  ?  Ans.  12  ft. 

2.  What  must  be  the  length  of  the  side  of  a  cubical  bin  that 
will  contain  150  bushels  ? 

3.  What  must  be  the  depth  of  a  cubical  cistern  that  will  hold 
200  bbl.  of  water  ? 

4.  How  many  sq.  ft.  in  the  surface  of  a  cube  whose  solidity  is 
79507  cu.  ft.  ?  Ans.  11094. 


404  SQUARE  AND  CUBE  ROOTS. 

PROBLEM  VI. 

694.  To  find  the  three  dimensions  of  a  parallelo- 
piped,  the  solid  contents  and  the  ratio  of  the  dimen- 
sions being  given. 

Note  1.  ^  This  case,  arithmetically  considered,  requires  us  to  separate  a  num- 
ber into  three  factors,  proportional  to  three  given  numbers. 

The  three  dimensions  will  be  like  multiples  of  the  proportional 
terms,  (691) ;  the  product  of  the  three  dimensions,  or  the  solid 
contents,  will  therefore  contain  the  product  of  the  three  propor- 
tional terms,  and  the  cube  of  the  common  ratio  which  the  pro- 
portional terms  respectively  bear  to  the  corresponding  dimensions, 
and  no  other  factor.     Hence  the 

EuLE.  I.  Divide  the  given  contents  hy  the  product  of  the  terms 
proportional  to  the  three  dimensions^  and  extract  the  cube  root  of 
the  quotient. 

II.  Multiply  the  root  thus  obtained  hy  each  proportional  term; 
the  products  will  he  the  corresponding  sides. 

Note  2.  —  The  dimensions  are  supposed  to  be  taken  in  a  direction  perpen- 
dicular to  the  faces  of  a  solid,  and  to  each  other. 

EXAMPLES   FOR  PRACTICE. 

1.  A  pile  of  bricks  in  the  form  of  a  parallelepiped  contains 
8000  cu.  ft ,  and  the  length,  breadth,  and  thickness,  are  to  ^ach 
other  as  4,  8,  and  2,  respectively;  what  are. the  dimensions  of  the 
pile?  Ans.  10,  15,  and  20  ft. 

2.  Three  numbers  are  to  each  other  as  2,  5,  and  7,  and  their 
continued  product  is  4480 ;  required  the  numbers. 

Ans.  8,  20,  and  28. 

8.  Separate  100  into  three  factors  which  shall  be  to  each  other 
as  2,  2 J,  and  8.       Ans.  3.76414  +  ;  4.70518  +  ;  5.64622—. 

4.  A  person  wishes  to  construct  a  bin  that  shall  be  of  equal 
width  and  depth,  and  the  length  three  times  the  width,  and  that 
shall  contain  450  bushels  of  grain  ?  what  must  be  its  dimensions  ? 


PROMISCUOUS  EXAMPLES.  405 

PROMISCUOUS    EXAMPLES. 

1.  There  is  a  park  containing  an  area  of  10  A.  2  R.  20  P.,  and 
the  breadth  is  equal  to  f  of  the  length.  If  two  men  start  from 
one  corner  and  travel  at  the  rate  of  3  miles  per  hour,  one  going 
by  the  walk  around  the  park,  and  the  other  taking  the  diagonal 
path  through  the  park,  how  much  sooner  will  the  latter  reach 
the  opposite  corner  than  the  former?  Ans.  1  min.  29.3  sec. 

2.  What  is  the  length  of  one  side  of  a  square  piece  of  land  con- 
taining 40  acres?  A7is.  80  rd. 

3.  The  ground  situated  between  two  parallel  streets  is  laid  out 
into  equal  rectangular  lots  whose  front  measure  is  44  per  cent, 
greater  than  the  depth.  Now,  if  the  streets  were  20  feet  further 
apart,  the  ground  could  be  laid  out  into  square  lots  of  the  same 
area  as  the  rectangular.    What  is  the  distance  between  the  streets  ? 

Ans.  100  feet. 

4.  How  much  less  will  it  cost  to  fence  40  acres  of  land  in  the 
form  of  a  square,  than  in  the  form  of  a  rectangle  of  which  the 
breadth  is  i  the  length,  the  price  per  rod  being  $1.40  ? 

Ans.  $112. 

5.  If  a  cistern  6  feet  in  diameter  holds  80  barrels  of  water,  how 
much  water  will  be  contained  in  a  cistern  of  the  same  depth  and 
1 8  feet  in  diameter  ? 

6.  What  is  the  length  of  the  side  of  a  square  which  contains 
the  same  area  as  a  rectangle  5i  by  7  feet  ?     Ans.  6  ft.  2.4  +  in. 

7.  What  is  the  length  of  the  side  of  a  square  which  can  just 
be  inclosed  within  a  circle  42  inches  in  diameter  ? 

Ans.  29.7  — in. 

8.  If  it  costs  $75  to  inclose  a  circular  fish  pond  containing  3  A. 
86  P.,  how  much  will  it  cost  to  inclose  another  containing  17  A. 
HOP.?  Ans.  $167.70. 

Note.  —  It  is  proved  in  Geometry  that  all  similar  solids  are  to  each  other  as 
the  cubes  of  their  like  dimensions.  Hence,  any  dimension  may  be  found  by 
proportion^  when  its  ratio  to  the  corresponding  dimension  of  a  known  similar 
solid  is  given. 

9.  What  is  the  length  of  the  side  of  a  cubical  vessel  that  con- 
tains J  as  much  as  one  whose  side  is  6  ft.  ?  Ans.  3  ft. 


406  SERIES. 

10.  How  many  globes  4  in.  in  diameter  are  equal  in  volume  to 
one.  12  in.  in  diameter? 

11.  If  an  ox  that  weighs  900  lb.  girt  6.5  ft.,  what  is  the  weight 
of  an  ox  that  girts  8  ft.  ?  Ans.  1677  lb.  14  -f  oz. 

12.  If  a  cable  3  in.  in  circumference  supports  a  weight  of  2500 
lb.,  what  must  be  the  circumference  of  a  cable  that  will  support 
4960  1b.? 

13.  If  a  stack  of  hay  4  feet  high  contain  4  tons,  how  high 
must  a  similar  stack  be  to  contain  20  tons  ? 


SERIES. 

G93.  A  Series  is  a  succession  of  numbers  so  related  to  each 
other,  that  each  number  in  the  succession  may  be  formed  in  the 
same  manner,  from  one  or  more  preceding  numbers.  Thus,  any 
number  in  the  succession,  2,  5,  8,  11,  14,  is  formed  by  adding  3 
to  the  preceding  number.     Hence,  2,  5,  8,  11,  14,  is  a  series. 

G90«  The  Law  of  a  Series  is  the  constant  relation  existing 
between  two  or  more  terms  of  the  series.  Thus,  in  the  series,  3, 
7,  11,  15,  we  observe  that  each  term  after  the  first  is  greater  than 
the  preceding  term  by  4;  this  constant  relation  between  the  terms 
is  the  law  of  this  series. 

The  law  of  a  series,  and  the  term  or  terms  on  which  it  de- 
pends being  given,  any  number  of  terms  of  the  series  can  be 
formed.  Thus,  let  64  be  a  term  of  a  series  whose  law  is,  that  each 
term  is  four  times  the  preceding  term.  The  term  following  64  is 
64  X  4,  the  next  term  64  x  4^,  etc.;  the  term  preceding  64  is 
64  -^-4.     Hence  the  series,  as  far  as  formed,  is  16,  64,  256, 1024. 

G97.  A  series  is  either  Ascending,  or  Descending^  according 
as  each  term  is  greater  or  less  than  the  preceding  term.  Thus,  2,  6, 
10, 14,  is  an  ascending  series;  32, 16,  8,  4,  is  a  descending  series. 

698.  An  Extreme  is  either  the  first  or  last  term  of  a  series. 
Thus,  in  the  series,  4,  7, 10, 13,  the  first  extreme  is  4,  the  last,  13. 

090.  A  Mean  is  any  term  between  the  two  extremes.  Thus, 
in  the  series,  5,  10,  20,  40,  80,  the  means  are  10,  20,  and  40. 


PROGRESSIONS.  407 

700.  An  Arithmetical  or  Eqnidifferent  Progression  is  a 

series  whose  law  of  formation  is  a  common  difference.  Thus,  in 
the  arithmetical  progression,  3,  7,  11,  15, 19,  each  term  is  formed 
from  the  preceding  by  adding  the  common  difference,  4. 

701.  An  arithmetical  progression  is  an  ascending  or  descend- 
ing series,  according  as  each  term  is  formed  from  the  preceding 
term  by  adding  or  subtracting  the  common  difference.  Thus,  the 
ascending  series,  7,  10,  13,  16,  etc.,  is  an  arithmetical  progression 
in  which  the  common  difference,  3,  is  constantly  added  to  form 
each  succeeding  term ;  and  the  descending  series,  20,  17,  14,  11, 
8,  5,  2,  is  an  arithmetical  progression  in  which  the  common  dif- 
ference is  constantly  subtracted,  to  form  each  succeeding  term. 

702.  A  Geometrical  Progression  is  a  series  whose  law  of 
formation  is  a  common  multiplier. .  Thus,  in  the  geometrical  pro- 
gression, 3,  6,  12,  24,  48,  each  term  is  formed  by  multiplying  the 
preceding  term  by  the  common  multiplier,  2. 

703*  A  geometrical  progression  is  an  ascending  or  descending 
series,  according  as  the  common  multiplier  is  a  whole  number  or 
a  fraction.  Thus,  the  ascending  series,  1,  2,  4,  8,  16,  etc.,  is  a 
geometrical  progression  in  which  the  common  multiplier  is  2- 
and  the  descending  series,  32,  16,  8,  4,  2,  1,  i,  J,  etc.,  is  a  geo- 
metrical progression  in  which  the  common  multiplier  is  J. 

704:.  The  Ratio  in  a  geometrical  progression  is  the  common 
multiplier. 

705.  In  the  solution  of  problems  in  Arithmetical  or  Geomet- 
rical progression,  five  parts  or  elements  are  concerned,  viz : 

In  Arithmetical  Progression  —        In  Geometrical  Progression  — 

1.  The  first  term  ;  1.  The  first  term  ; 

2.  "  last  term;  2.  "  last  term ; 

3.  "  number  of  terms ;  3.  **  number  of  terms ; 

4.  "  common  difference ;  4.  "  ratio ; 

5.  *'  sum  of  the  series.  5.  "  sum  of  the  series. 

The  conditions  of  a  problem  in  progression  may  be  such  as  to 
require  any  one  of  the  five  parts  from  any  three  of  the  four  re- 
maining parts ;  hence,  in  either  Arithmetical  or  Geometrical  Pro- 
gression, there  are  5  x  4  =  20  cases,  or  classes  of  problems,  and 
no  more,  requiring  each  a  different  solution. 


408  SERIES. 

GENERAL  PROBLEMS  IN  ARITHMETICAL  PROGRESSION. 
PROBLEM    I. 

706.  Given,  one  of  the  extremes,  the  common  dif- 
ference, and  the  number  of  terms,  to  find  the  other 
extreme. 

Let  2  be  the  first  term  of  an  arithmetical  progression,  and  3  the 
common  difference ;  then, 

2  ==2  =2,  1st  term. 

2+3  =-  2  +  (3  X  1)  =   5,  2d      " 

2  +  3  +  3  --2+(3  X  2)=  8,  3d  " 
2  +  3  +  3  +  3  =  2+ (3  X  3)  =  11,  4th  " 
From  this  illustration  we  perceive  that,  in  an  arithmetical  pro- 
gression, when  the  series  is  ascending,  the  second  term  is  equal  to  the 
first  term  plus  the  common  difference  ;  the  third  term  is  equal  to  the 
first  term  plus  2  times  the  common  difference ;  the  fourth  term  is 
equal  to  the  first  term  plus  3  times  the  common  difference ;  and  so  on. 
In  a  descending  series,  the  second  term  is  equal  to  the  first  term 
minus  the  common  difference ;  the  third  term  is  equal  to  the  first 
minus  2  times  the  common  difference ;  and  so  on.  In  all  cases  the 
difference  between  the  two  extremes  is  equal  to  the  product  of  the 
common  difference  by  the  number  of  terms  less  1.     Hence  the 

Rule.  Multiply  the  common  difference  hy  the  number  of  terms 
less  1  /  add  the  product  to  the  given  term  if  it  he  the  less  extreme^ 
and  subtract  the  product  from  the  given  term  if  it  be  the  greater 
extreme. 

EXAMPLES   FOR   PRACTICE. 

1.  The. first  term  of  an  arithmetical  progression  is  5,  the  com- 
mon difference  4,  and  the  number  of  terms  8 ;  what  is  the  last 
term?  Ans.  33. 

2.  If  the  first  term  of  an  ascending  series  be  2,  and  the  com- 
mon difference  3,  what  is  the  50th  term  ? 

3.  The  first  term  of  a  descending  series  is  100,  the  common 
difference  7,  and  the  number  of  terms  13  ;  what  is  the  last  term  ? 

4.  If  the  first  term  of  an  ascending  series  be  |,  the  common 
difference  f ,  and  the  number  of  terms  20,  what  is  the  last  term  ? 

Ans.  1^1, 


ARITHMETICAL  PROGRESSION.  40g 

PROBLEM   II. 

7®7.  Given,  the  extremes  and  number  of  terms,  to 
find  the  common  difference. 

Since  the  difference  of  the  extremes  is  always  equal  to  the  common, 
difference  multiplied  by  the  number  of  terms  less  1,  (706),  we  have 
the  following  ^ 

Rule.  Divide  the  difference  of  (he  extremes  h^tJie  number  of 
terms  less  1. 

EXAMPLES   FOR   PRACTICE. 

1.  If  the  extremes  of  an  arithmetical  series  are  3  and  15,  and 
the  number  of  terms  7,  what  is  the  common  difference  ? 

Ans.  2. 

2.  The  extremes  are  1  and  51,  and  the  number  of  terms  is  76; 
what  is  the  common  difference  ? 

3.  The  extremes  are  .05  and  .1,  and  tlie  number  of  terms  is  8 ; 
what  is  the  common  difference?  Ans,  .00714285. 

4.  If  the  extremes  are  0  and  2  J,  and  the  number  of  terms  is 
18,  what  is  the  common  difference  ? 

PROBLEM    HI. 

708.  Given,  the  extremes  and  common  difference, 
to  find  the  number  of  terms. 

Since  the  difference  of  the  extremes  is  equal  to  the  common  differ- 
ence multiplied  by  the  number  of  terms  less  1,  (706),  we  have  tho 
following 

Rule.  Divide  the  difference  of  the  extremes  hy  the  common 
difference,  and  add  1  to  the  quotient. 

EXAMPLES    FOR   PRACTICE. 

1.  The  extremes  of  an  arithmetical  series  are  5  and  75,  and  the 
common  difference  is  5 ;  what  is  the  number  of  terms  ? 

Ana.   15. 

2.  The  extremes  are  J  and  20,  and  the  common  difference  is 
G} ;  find  the  number  of  terms. 

35 


410  SEKIES. 

3.  The  extremes  are  2.5  and  .25,  and  the  common  difference  is 
.125;   what  is  the  number  of  terms? 

4.  Insert  5  arithmetical  means  between  2  and  37. 

PROBLEM  IV.       ^        /  ,/-  ^     , 

/  709w  Given,  Ijie  extremes  and  nuniber  of  terms,  to 

iind  the  sum  ol  the  series.        # 

Let  VIS  take  any  series,  as  2,  5,  8,  11,  14,  and  writing  under  it  the 
same  series  in  an  inverse  order,  add  each  term  of  the  inverted  series 
to  the  term  above  it  in  the  direct  series,  thus : 

2+5+    8+11  +  14  =  40,  once  the  sum, 
14 -^11 -J-    8+    5+    2  =  40,      "      ''      *' 

IG  +  IG  +  IG  +  16  +  16  =  80,  twice  the  sum. 

From  this  we  perceive  that  16,  the  sum  of  the  extremes  of  the  given 
series,  multiplied  by  5,  the  number  of  terms,  equals  80,  w^hich  is  tivice 
the  sum  of  the  series  ;  and  80  -7-  2  =  40,  the  sum  of  the  series.    Hence 

EuLE.  Mulliphj  the  sum  of  the  extremes  hy  the  number  of 
terms,  and  divide  the  product  hi/  2.  a/yi,.<X#t  ^f/ 

J"^'     M.       \»      L^«-^  EXAMPLES   FOlf  PRACTICE. 

.   1.  Find  the  sum  of  the  series  the  first  term  of  which  is  4,  the 
common  difference  6,  and  the  last  term  40.  Ans.  154. 

2.  The  extremes  are  0  and  250,  and  the  number  of  terms  is 
ICOO ;  w^hat  is  the  sum  of  the  series  ? 

3.  A  person  wishes  to  discharge  a  debt  in  11  annual  payments 
such  that  the  last  payment  shall  be  $220,  and  each  payment  greater 
than  the  preceding  by  §17;  find  the  amount  of  the  debt,  and  the 
first  payment.  Ans.   First  payment,  $50. 

TIO.  By  reversing  some  one  of  the  four  pr^lems  now  given, 
or  by  combining  two  or  more  of  them,  all  of  the  sixteen  remain- 
ing problems  of  Arithmetical  Progression  may  be  solved  or 
analyzed. 


GEOMETRICAL   PROGRESSION.  411 

GENERAL  PROBLEMS  IN  GEOMETRICAL  PROGRESSION. 
PROBLEM    I. 

711.  Given,  one  of  the  extremes,  the  ratio,  and  the 
number  of  terms,  to  find  the  other  extreme. 

Let  3  be  the  first  term  of  a  geometrical  progression,  and  2  tho 
ratio:  then, 

3  =3=3,  the  1st  term, 

3x2  =  3  X  21  =    6,  *'  2d      " 

3x2x2         =  3  X  22  =  12,  ''3d      " 

3  X  2  X  2  X  2  =  3  X  23  =  24,  "4th     " 

From  this  illustration  we  perceive  that,  in  a  geometrical  progression, 
the  second  term  is  equal  to  the  first  term  multiplied  by  the  ratio ;  the 
third  term  is  equal  to  the  first  term  multiplied  by  the  second  power 
of  the  ratio ;  the  fourth  term  is  equal  to  the  first  term  multiplied  by 
the  third  power  of  the  ratio ;  and  so  on.  The  same  is  true  whether 
the  ratio  be  an  Integer  or  fraction.     Hence  the  following 

HuLE.  I.  If  the  given  extreme  he  the  first  terrrij  multiply  it  hy 
that  power  of  the  ratio  indicated  hy  the  number  of  terms  less  1 ; 
the  result  icill  he  the  last  term. 

II.  If  the  given  extreme  he  the  last  term^  divide  it  hy  that  power 
of  the  ratio  indicated  hy  the  number  of  terms  less  1 ;  the  result 
will  be  the  first  term. 

EXAMPLES   FOR   PRACTICE. 

1.  The  first  term  of  a  geometrical  series  is  6,  the  ratio  4,  and 
the  number  of  terms  6 ;  find  the  last  term.  Ans.  6144. 

2.  The  last  term  of  a  geometrical  series  is  192,  the  ratio  2,  and 
the  number  of  terms  7 ;  what  is  the  first  term  ? 

3.  If  the  first  term  be  6,  the  ratio  ^,  and  the  number  of  terms 
8,  what  is  the  last  term  ? 

4.  The  first  term  is  25,  the  ratio  |,  and  the  number  of  terms 
5 ;  what  is  the  last  term  ?  Ans.  ■^~^, 


412  SERIES. 

PROBLEM   II. 

TIS.  Given,  the  extremes  and  number  of  terms,  to 
find  the  ratio. 

Since  the  last  term  is  always  equal  to  the  first  term  multiplied  by 
that  power  of  the  ratio  indicated  by  the  number  of  terms  less  1, 
(711),  we  have  the  following 

Rule.  Divide  the  last  term  hy  the  fir  sty  and  extract  that  root 
of  the  quotient  indicated  hy  the  number  of  terms  less  1 ;  the  result 
will  be  the  ratio, 

EXAMPLES   FOR   PRACTICE. 

1.  The  extremes  are  2  and  512,  and  the  number  of  terms  is  5 ', 
what  is  the  ratio  ?  Ans.  4. 

2.  The  extremes  are  ^^  and  45 y^^,  and  the  number  of  terms  is 
8;  what  is  the  ratio? 

8.  The  extremes  are  7  and  .0112,  and  the  number  of  terms  is 
5 ;  what  is  the  ratio  ?  Aiis.  5. 

4.  Insert  3  geometrical  means  between  8  and  5000. 

PROBLEM    III. 

713.  Given,  the  extremes  and  ratio,  to  find  the  num- 
ber of  terms. 

Since  the  quotient  of  the  last  term  divided  by  the  first  term  is 
equal  to  that  power  of  the  ratio  indicated  by  the  number  of  terms 
less  1,  (712),  we  have  the  following 

IluLE.  Divide  the  last  term  by  the  first y  divide  this  quotient  hy 
the  ratioj  and  the  quotient  thus  obtained  hy  the  ratio  again,  and  so 
on  in  successive  division,  till  the  final  quotient  ts  1.  The  number 
of  times  the  ratio  is  used  as  a  divisor,  plus  1,  is  the  number  of 
terms. 

examples;  for  practice. 

1.  The  extremes  are  2  and  1458,  and  the  ratio  is  3 ;  what  is 
the  number  of  terms  ?  Ans.  7. 

2.  The  first  term  is  .1,  the  last  term  100,  and  the  ratio  10 ;  find 
the  number  of  terms. 


GEOMETRICAL  PROGRESSION.  413 

3.  The  first  term  is  g|^,  the  last  term  i,  and  the  ratio  2;  what 
is  the  number  of  terms  ? 

4.  The  extremes  are  196608  and  6,  and  the  ratio  is  i ;  what  is 
the  number  of  terms  ?  Ans.  Q. 

PROBLEM    IV. 

714.  Given,  the  extremes  and  ratio,  to  find  tlie  sum 
of  the  series. 

Let  us  take  the  series  5  +  20  4-  80  +  820=^425,  multiply  each  term 
by  the  ratio  4,  and  from  this  result  subtract  the  given  series  term  from 
term,  thus : 

20  +  80  +    320  +  1280  =  1700,  four  times  the  series, 

5  4-  20  +  80  4-    320  =  425,  once  the  series, 

1280  —    5     =^  1275,  three  times  the  series, 

Then  1275  ~    3     =    425,  once  the  series. 

Hence  the 

Rule.   Multiply  the  greater  extreme  hy  the  ratioj  suhtract  tlie 

less  extreme  from  the  product^  and  divide  the  remainder  by  the 

ratio  less  1. 

NoTK. — Let  every  descending  series  be  inverted,  and  the  first  term  called  the 
last ;  then  the  ratio  will  be  greater  than  a  unit.  If  the  series  be  infinite^  the  least 
term  is  a  cipher. 

EXAMPLES    FOR   PRACTICE. 

1.  The  extremes  are  3  and  384,  and  the  ratio  is  2;  what  is  the 
Bum  of  the  series  ?  Ans,  765. 

2.  If  the  extremes  are  5  and  1080,  and  the  ratio  is  6,  what  is 
the  sum  of  the  series  ? 

3.  If  the  first  term  is  4|,  the  last  term  ^|^,  and  the  ratio  i, 
what  is  the  sum  of  the  series  ?  Ans,  7£-J^. 

4.  What  is  the  sum  of  the  infinite  series,  8,  4,  2,  1,  i,  J,  etc.? 

PROBLEM    V. 

715.  Given,  the  first  term,  the  ratio,  and  the  num- 
ber of  terms,  to  find  the  sum  of  the  series. 

If,  for  example,  the  first  term  be  4,  the  ratio  3,  and  the  number  of 
terms  6,  then  by  Problem  I,  we  have 

4x3^=^  the  last  term. 
35* 


414  SERIES. 

Whence  by  Problem  IV,  we  nave 

— —  =  ^—- - — =  1456,  the  sum  of  the  series, 

o  —  1  o  —  1 

Hence  the  following 

KuLE.  Raise  the  ratio  to  a  power  indicated  hy  the  number  of 
terms,  and  subtract  1  from  the  result ;  then  multiply  this  remainder 
by  the  first  term^  and  divide  the  product  by  the  ratio  less  1. 

EXAMPLES   FOR   PRACTICE. 

1.  The  first  term  is  7,  the  ratio  3,  and  the  number  of  terms  4 ; 
what  is  the  sum  of  the  series  ?  Ans.  280. 

2.  The  first  term  is  375,  the  ratio  ^,  and  the  number  of  terms 
4 ;  what  is  the  sum  of  the  series  ? 

3.  The  first  term  is  175,  the  ratio  1.06,  and  the  number  of  terms 
5;  what  is  the  sum  of  the  series?  Ans,  986.49-[-. 

PROBLEM   VL 

716.  Given,  the  extremes  and  the  sum  of  the  series, 
to  find  the  ratio. 

If  we  take  the  geometrical  progression,  2,  6,  18,  54,  162,  in  which 
the  ratio  is  3,  and  remove  the  first  term  and  the  last  term,  succes- 
sively, and  then  compare  the  results,  we  have 

6  +  18  4-  54  +  162  =  sum  of  the  series  minus  the  first  term. 

2  4-    6  4-  18  4-    54  =  sum  of  the  series  minus  the  last  term. 
Now,  since  every  term  in  the  first  line  is  3  times  the  corresponding 
term  in  the  second  line,  the  sum  of  the  terms  in  the  first  line  must 
be  3  times  the  sum  of  the  terms  in  the  second  line.     Hence  the 

Rule.  Divide  the  sum  of  the  series  minus  the  first  term,  by  the 
sum  of  the  series  minus  the  last  term, 

EXAMPLES   POR   PRACTICE. 

1.  The  extremes  are  2  and  686,  and  the  sum  of  the  series  is 
800 ;  what  is  the  ratio  ?  Ans.  7. 


GEOMETRICAL  PROGRESSIOX.  415 

2.  The  extremes  are  J  and  G4,  and  the  sum  of  the  scries  is 
127|;  what  is  the  ratio? 

3.  If*  the  sum  of  an  infinite  series  be  42,  and  the  greater  ex- 
treme 3,  what  is  the  ratio  ?  Ans.  J. 

TIT'.  Every  other  problem  in  Geometrical  Progression,  that 
admits  of  an  arithmetical  solution,  may  be  solved  either  by  re- 
versing or  combining  some  of  the  problems  already  given. 

COMPOUND    INTEREST    BY    GEOxMETRlCAL    PROGRESSION. 

71§,  We  have  seen  (^^clO)  that  if  any  sum  at  compound  in- 
terest be  multiplied  by  the  amount  of  81  ibr  the  given  interval, 
the  product  will  be  the  amount  of  the  given  sum  or  principal  at 
the  end  of  the  first  interval;  and  that  this  amount  constitutes  a 
new  principal  for  the  second  interval,  and  so  on  for  a  third,  fourth, 
or  any  other  interv^al.     Hence, 

A  question  in  compound  interest  constitutes  a  geometrical  pro- 
gression, whose  first  term  is  the  principal ;  the  common  multiplier 
or  ratio  is  one  plus  the  rate  per  cent,  for  one  interval;  the  number 
of  terms  is  equal  to  the  number  of  intervals  -j-1 ;  and  the  last 
term  is  the  amount  of  the  given  principal  for  the  given  time.  All 
the  usual  cases  of  compound  interest  and  discount  computed  at 
compound  interest,  can  therefore  be  solved  by  the  rules  for  geo- 
metrical progression.     For  example, 

Find  the  amount  of  $250  for  4  years,  at  6  %  compound 
interest. 

OPERATION". 

$250  X  1.06^  =  S250  X  1.262477  «=  S316.21925. 

Analysis.  Here  we  have  $250  the  first  term,  1.06  the  ratio,  and 
5  the  number  of  terms,  to  find  the  last  term.  Then  by  711  we  find 
the  last  term,  which  is  the  amount  required. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  amount  of  $350  in  4  years,  at  6  %  per  annum 
compound  interest  ?  Ans.  8441.86. 

2.  Of  what  principal  is  §150  the  compound  interest  fui  2  years, 
at  7  %  ? 


416  SERIES. 

3.  What  sum  at  6  %  compound  interest,  will  amount  to  $1000 
in  3  years  ?  Ans.   $839.62. 

4.  In  how  many  years  will  $40  amount  to  $53.24,  at  10  %  com- 
pound interest?  Ans.  8  years. 

5.  At  what  rate  per  cent,  compound  interest  will  any  sum  double 
itself  in  8  years?  Ans.  9.05  +  %. 

6.  What  is  the  present  worth  of  $322.51,  at  5  %  compound 
interest,  due  24  years  hence?  Ans,  $100. 

>'>-^  ANNUITIES.  ^  — 


0^S^ 


fi*-  -+- 


719*  An  Annuity  is  literally  a  sum  of  money  which  is  pay- 
able annually.  The  term  is,  however,  applied  to  a  sum  which  is 
payable  at  any  equal  intervals,  as  monthly,  quarterly,  semi-annu- 
ally, etc. 

Note. — The  term,  interval,  will  be  used  to  denote  the  time  between  payments. 

Annuities  are  of  three  kinds :  Certain,  Contingent,  and  Per- 
petual. 

yS©.  A  Certain  Annuity  is  one  whose  period  of  continu- 
ance is  definite  or  fixed. 

ySfl.  A  Contingent  Annuity  is  one  whose  time  of  commence- 
ment, or  ending,  or  both,  is  uncertain ;  and  hence  the  period  of 
its  continuance  is  uncertain. 

722.  A  Perpetual  Annuity  or  Perpetuity  is  one  which  con- 
tinues forever. 

723.  Each  of  these  kinds  is  subject,  in  reference  to  its  com- 
mencement, to  the  three  following  conditions  : 

1st.  It  may  he  deferred^  i.  e.,  it  is  not  to  be  entered  upon  until 
after  a  certain  period  of  time. 

2d.  It  may  he  reversionary^  i.  e.,  it  is  not  to  be  entered  upon 
until  after  the  death  of  a  certain  person,  or  the  occurrence  of  some 
certain  event. 

Sd.  It  may  he  in  possession^  i.  e.,  it  is  to  be  entered  upon  at 
once. 


ANNUITIES.  417 

724.  An  Annuity  in  Arrears  or  Forborne  is  one  on  which 
the  payments  were  not  made  when  due.  Interest  is  to  be  reck- 
oned on  each  payment  of  an  annuity  in  arrears,  from  its  maturity, 
the  same  as  on  any  other  debt. 

ANNUITIES  AT  SIMPLE  INTEREST. 

725.  In  reference  to  an  annuity  at  simple  interest,  we  observe : 

I.  The  first  payment  becomes  due  at  the  end  of  the  first  inter- 
val, and  hence  will  bear  interest  until  the  annuity  is  settled. 

II.  The  second  payment  becomes  due  at  the  end  of  the  second 
interval,  and  hence  will  bear  interest  for  one  interval  less  than  the 
first  payment. 

III.  The  third  payment  will  bear  interest  for  one  interval  less 
than  the  second;  and  so  on  to  any  number  of  terms.     Hence, 

TV.  All  the  payments  being  settled  at  one  time,  each  will  be 
less  than  the  preceding,  by  the  interest  on  the  annuity  for  one 
interval.  Therefore,  they  will  constitute  a  descending  arithmetical 
progression,  whose  first  term  is  the  annuity  plus  its  interest  for  as 
many  intervals  less  one  as  intervene  between  the  commencement 
and  settlement  of  the  annuity;  the  common  difference  is  the  in- 
terest on  the  annuity  for  one  interval ;  the  number  of  terms  is  the 
number  of  intervals  between  the  commencement  and  settlement^ 
of  the  annuity;  and  the  last  term  is  the  annuity  itself. 

726.  The  rules  in  Arithmetical  Progression  will  solve  all 
problems  in  annuities  at  simple  interest. 

EXAMPLES    FOR    PRACTICE. 

1.  A  man  works  for  a  farmer  one  year  and  six  months,  at  $20 
per  month,  payable  monthly;  and  these  wages  remain  unpaid 
until  the  expiration  of  the  whole  term  of  service.  How  much  is 
due  to  the  workman,  allowing  simple  interest  at  6  per  cent,  per 
annum  ? 

OPERATION.  '  -  Analysis.    Here  the 

S20  +  UO  X  17  =  $21.70,  first  term,     l^^^    month's     wages, 
^20  +  $21.70  ^^^'  ^^  *^^  ^^^^  term; 

^ X  18  =  375.30,  sum.         the  number  of  months, 

18,  is  the   number  of 
2b 


418  SERIES. 

terms ;  and  the  interest  on  1  month^s  wages,  $.10,  is  the  common  dif- 
ference ;  and  since  the  first  month's  wages  has  been  on  interest  17 
months,  the  progression  is  a  descending  series.  Then,  by  706  we  find 
the  first  term,  which  is  the  amount  of  the  first  month's  wages  for  17 
months ;  and  by  709  we  find  the  sum  of  the  series,  which  is  the  sum 
of  all  the  wages  and  interest. 

2.  A  father  deposits  annually  for  the  benefit  of  his  son,  com- 
mencing with  his  tenth  birthday,  such  a  sum  that  on  his  21st 
birthday  the  first  deposit  at  simple  interest  amounts  to  $210,  and 
the  sum  due  his  son  to  $1860.  How  much  is  the  deposit,  and  at 
what  rate  per  cent,  is  it  deposited  ? 

OPERATION.  Analysis.  Here  the 

$1860x2— $210x12       ^-.^^    J         .,  $210,  the  amount  of 

Y^ =  ^1^0,  deposit.        the   first   deposit,    is 

21Q IQO  t^6   fi^st    term  ;    12, 

-jj" =  10  %,  rate.  the  number  of  depo- 
sits, is  the  number  of 
terms ;  and  $1860,  the  amount  of  all  the  deposits  and  interests,  is  the 
sum  of  the  series.  By  709  we  find  the  last  term  to  be  $100,  which 
is  the  annual  deposit ;  and  by  707  we  find  the  common  difi'erence  to 
be  $10,  which  is  the  annual  rate  % . 

3.  What  is  the  amount  of  an  annuity  of  $150  for  o  J  years,  pay- 
able quarterly,  at  IJ  per  cent,  per  quarter?         Aiis.  $3819.75. 

4.  In  what  time  will  an  annual  pension  of  $500  amount  to 
$3450,  at  6  per  cent,  simple  interest  ?  Ans.  6  years. 

5.  Find  the  rate  per  cent  at  which  an  annuity  of  $6000  will 
amount  to  $59760  in  8  years,  at  simple  interest. 

Ans.  7  per  cent. 

ANNUITIES  AT  COMPOUND  INTEREST. 

727.  An  Annuity  at  compound  interest  constitutes  a  geomet- 
rical progression  whose  first  term  is  the  annuity  itself;  the  common 
multiplier  is  one  plus  the  rate  per  cent,  for  one  interval  expressed 
decimally  5  the  number  of  terms  is  the  number  of  intervals  for  which 
the  annuity  is  taken;  and  the  last  term  is  the  first  term  multiplied 
by  one  plus  the  rate  per  cent,  for  one  interval  raised  to  a  power 
one  less  than  the  number  of  terms. 


PPtOMISCUOUS  EXAMPLES.  4I9 

73S.  The  Present  Value  of  an  Annuity  is  sucli  a  sum  as 
would  produce,  at  compound  interest,  at  a  given  rate,  the  same 
amount  as  the  sum  of  all  the  payments  of  the  annuity  at  com- 
pound interest.  Hence,  to  find  the  present  value; — First  find  the 
amount  of  the  annuity  at  the  given  rate  and  for  the  given  time  hy 
K^5\  the/ii  find  the  present  value  of  this  amount  h[^  5*^^^ 
talcing  out  the  anioimt  o/Sl,  or  divisor^  from  ^•51, 

Notes. — 1.  The  present  value  of  a  rcverpionary  annuity  is  that  principal  which 
will  amount,  at  the  time  the  reverb-ion  expiree,  to  what  will  theu  be  the  prOi>ei)t 
value  of  the  annuity. 

2.  The  present  value  of  a  perpetuity  is  a  sum  whose  interest  equals  the  an- 
nuity. 

"^SO.  Questions  in  Annuities  at  compound  interest  can  be 
solved  by  the  rules  of  Geometrical  Progression. 


PROMISCUOUS    EXAMPLES    1^    SERIES. 

1.  Allowing  G  per  cent,  compound  interest  on  an  annuity  of 
$200  which  is  in  arrears  20  years^  what  is  its  present  amount  ? 

Ans.  $7857.11. 

2.  Find  the  annuity  whose  amount  for  25  years  is  81G459.85, 
allowing  compound  interest  at  6  per  cent.  Ans.   $300. 

3.  What  is  the  present  worth  of  an  annuity  of  $500  for  7  years, 
at  6  per  cent,  compound  interest?  Ans.  82791.18. 

4.  What  is  the  present  value  of  a  reversionary  lease  of  $100, 
commencing  14  years  hence,  and  to  continue  20  years,  coujpound 
interest  at  5  per  cent.?  Ans.  §629.420. 

5.  Find  the  sum  of  21  terms  of  the  series,  5,  4|,  4  J,  etc. 

6.  A  man  traveled  13  days;  his  last  day's  journey  was  80  miles, 
and  each  day  he  traveled  5  miles  more  than  on  the  preceding  day. 
How  far  did  he  travel,  and  what  was  his  first  day's  journey? 

Ans.   He  traveled  650  miles. 

7.  Find  the  12th  term  of  the  scries,  30,  15,  7  J,  etc. 

8.  The  first  term  of  a  geometrical  progression  is  2,  the  last  term 
512,  and  common  multiplier  4;  find  the  sum  of  the  scries. 

Ans.  682. 


420  SEllIES. 

9.  The  distance  between  two  places  is  360  miles.  In  liow  many 
days  can  it  be  traveled,  by  a  man  who  travels  the  first  day  27 
miles,  and  the  last  day  45,  each  day's  journey  being  greater  than 
the  preceding  by  the  same  number  of  miles  ?  Ans.  10. 

10.  The  first  term  of  a  geometrical  progression  is  1,  the  last 
term  15625,  and  the  number  of  terms  7;  find  the  common  ratio. 

Ans.  5. 

11.  An  annual  pension  of  S500  is  in  arrears  10  years.  What 
is  the  amount  now  due,  allowing  6  per  cent,  compound  interest  ? 

Ans,  $6590.40. 
/  12.   Find  the  first  and  last  terms  of  an  arithmetical  progression 
whose  sum  is  408,  common  difi*erence  6,  and  number  of  terms  8.  - 

Ans.  First  term,  30 ;  last  term,  72. 

13.  A  farmer  pays  $1196,  in  13  quarterly  payments,  in  such  a 
way  that  each  payment  is  greater  than  the  preceding  by  $12. 
What  are  his  first  and  last  payments  ?  Ans.  $20,  and  $164. 

14.  A  man  wishes  to  discharge  a  debt  in  yeany  payments,  mak- 
ing the  first  payment  $2,  the  hist  $512,  and  each  payment  four 
times  the  preceding  payment.  How  long  will  it  take  him  to  dis- 
charge the  debt,  and  what  is  the  amount  of  his  indebtedness  ? 

15.  A  man  dying,  left  5  sons,  to  whom  he  gave  his  property  as 
follows :  to  the  youngest  he  gave  §4800,  and  to  each  of  the  others 
1 J  times  the  next  younger  son's  share.  What  was  the  eldest  son's 
fortune,  and  what  the  amount  of  property  left  ? 

Ans.  Eldest  son's  share,  $24300;  property,  $63300. 

16.  Find  the  annuity  whose  amount  for  5  years,  at  6  per  cent, 
compound  interest,  is  $2818.546.  Ans.  $500. 

17.  A  merchant  pays  a  debt  in  yearly  payments  in  such  a  way 
that  each  payment  is  3  times  the  preceding;  his  first  payment  is 
$10,  and  his  last  $7290.  What  is  the  amount  of  the  debt,  and  in 
how  many  payments  is  it  discharged  ? 

Ans.  Debt,  $10930;  7  payments. 

18.  A  man  traveling  along  a  road,  stopped  at  a  numl)er  of 
stations,  but  at  each  station  he  found  it  necessary,  before  proceed- 
ing to  the  next,  to  return  to  the  place  from   which  he  first  started  ; 


I 


PROMISCUOUS  EXAMPLES.  421 


the  distance  from  the  starting  place  to  the  first  station  was  5  miles, 
and  to  the  last  25  miles;  he  traveled  in  all  180  miles.  How 
many  stations  were  there  on  the  road,  and  what  was  the  distance 
from  station  to  station  ?  Ans.  6  stations ;  4  miles  apart. 

19.  An  annuity  of  $200  for  12  years  is  in  reversion  6  years 
"What  is  its  present  worth,  compound  interest  at  6  %  ? 

Ans,UlS2M  +  . 

20.  A  man  pays  $6  yearly  for  tobacco,  from  the  age  of  16  until 
he  is  60,  when  he  dies,  leaving  to  his  heirs  $500.  What  might 
he  have  left  them,  if  he  had  dispensed  with  this  useless  habit  and 
loaned  the  money  at  the  end  of  each  year  at  6  %  compound 
interest?  Ans,  $1698.548+. 

21.  What  is  the  present  worth  of  a  reversionary  perpetuity  of 
$100,  commencing  30  years  h(^nce,  allowing  5  per  cent,  compound 
interest?  Ans.  $462.75+. 

22.  Two  boys,  each  12  years  old,  have  certain  sums  of  money 
left  to  them ;  the  sum  left  to  one  is  put  out  at  7  %  simple  inte- 
rest, and  the  sum  left  the  other  at  6  %  compound  interest,  paya- 
ble semi-annually,  and  the  amount  of  each  boy's  money  will  be 
$2000  when  he  is  21  years  old.  What  is  the  sum  left  Jo  each 
boy? 

23.  A  merchant  purchased  8  pieces  of  cloth,  for  which  he  paid 
$136;  the  difference  in  the  length  of  any  two  pieces  was  2  yds. 
and  the  difference  in  the  price  $4.  He  paid  $31  for  the  longest 
piece,  and  $1  a  yard  for  the  shortest.  Find  the  whole  number  of 
yards,  and  the  price  per  yard  of  each  piece. 

24.  A  farmer  has  600  bushels  of  different  kinds  of  grain,  mixed 
in  such  a  way  that  the  number  of  bushels  of  the  several  kinds  con- 
stitute a  geometrical  progression,  whose  common  multiplier  is  2 ; 
the  greatest  number  of  bushels  of  one  kind  is  320.  Find  the 
number  of  kinds  of  grain  in  the  mixture,  and  the  number  of 
bushels  of  each  kind.  Ans.  4  kinds. 

86 


422  MISCELLANEOUS   EXAMPLES. 


MISCELLANEOUS  EXAMPLES. 

1.  How  many  thousand  shingles  will  cover  both  sides  of  a  roof  36 
ft.  long,  and  wiiose  rafters  are  18  ft.  in  length  ? 

2.  From  f  of  -^  of  i  of  70  miles,  subtract  .73  of  1  mi.  3  fur. 

3.  What  number  is  that  from  which  if  7J  be  subtracted,  f  of  the 
remainder  is  91^?  Ans.  144 J. 

4.  What  part  of  4  is  |  of  6?  Ans.  |. 

5  It  is  required  to  mix  together  brandy  at  $.80  a  gallon,  wine  at 
$.70,  cider  at  $.10,  and  water,  in  such  proportions  that  the  mixture 
may  be  worth  $.50  a  gallon;  what  quantity  of  each  must  be  used? 

Ans,  3  gal.  of  water,  2  of  cider,  4  of  wine,  and  5  of  brandy. 

6.  What  number  increased  by  J,  i,  and  J  of  itself  equals  125  ? 

7.  What  is  the  hour,  when  the  time  past  uoon  is  equal  to  f  of  the 
time  to  midnight?  Ans.  4  h.  48  min.  p.  m. 

8.  A  grocer  mixed  12  cwt.  of  sugar  @  $10,  with  3  cwt.  @  $8f ,  and 
8  cwt.  @  $7i;  how  much  was  1  cwt.  of  the  mixture  worth? 

9.  If  $240  gain  $5.84  in  4  mo.  2G  da.,  what  is  the  rate  fo  ?    J^ns.  6. 

10.  If  24  men,  in  189  da.,  working  10  h.  a  day,  dig  a  trench  33 J  yd, 
long,  2|  yd.  deep,  and  5j-  yd.  wide;  how  many  hours  a  day  must  217 
men  work,  to  dig  a  trench  23 i  yd.  long,  2 J  yd.  deep,  and  3|  yd.  wide, 
in  5}  days?  Ans.  10  h. 

11.  What  is  the  difference  between  the  interest  and  the  discount  of 
$450  at  5  per  cent.,  for  6  jr.  10  mo.? 

12.  A  younger  brother  received  $G300,  which  was  i  as  much  as  his 
elder  brother  received;  how  much  did  both  receive? 

13.  Reduce  .7,  .88,  .727,  .91325  to  their  equivalent  common  fractions. 

14.  A  person  by  selling  a  lot  of  goods  for  $438,  loses  10  ^c ;  how 
much  should  the  goods  have  been  sold  for,  to  gain  12J  ^? 

15.  For  what  sum  must  a  note  be  drawn  at  4  mo.,  that  the  proceeds     ^ 
of  it,  when  discounted  at  bank  at  7  per  cent.,  shall  be  $875.50?    ^9  ^^/ 

16.  Three  persons  engaged  in  trade  with  a  joint  capital  of  $2128; 
A's  capital  was  in  trade  5  mo.,  B's  8  mo.,  and  C's  12  mo.;  A's  share 
of  the  2;ain  was  $228,  B's  $266.40,  and  C's  $330.  What  was  the  capital 
of  each?  Ans.  A's,  $912  ;  B's,  $666;  C's,  ^^555. 


crG 

Sep..  ., —  ^__,  ^..  _  ,  ^  J,.         ,  ^  .. 

@  $.40.    When  was  the  a|c  due  per  average?  Ans.  iNov.  8. 

18.  A  B  and  C  can  do  a  job  of  work  in  12  da.,  C  can  do  it  in  24  da., 
and  A  in  34  da. ;  in  what  time  can  B  do  it  alone?        Ans.  81^  da. 

19.  If  a  man  travel  7  mi.  the  first  day,  and  51  mi.  the  last,  increas- 
ing his  journey  4  mi.  each  day,  how  many  days  will  he  travel,  and 
how  far?  Ans.  12  da.,  and  348  mi. 


MISCELLANEOUS    EXAMPLES.  423 

20.  What  is  the  difference  between  the  true  and  bank  discount  of 
$2500,  payable  in  90  days  at  7  per  cent.  ?  Ans.  5;>2.21. 

21.  Which  is  the  more  advantageous,  to  buy  flour  at  $5  a  bbl.  on  6 
mo.,  or  $4.87  J  cash,  money  being  worth  7  ^  ?      Ans.  At  $5  on  G  mo. 

22.  Sold  J  of  a  lot  of  lumber  for  what  |  of  it  cost ;  what  fc  was 
gained  on  the  part  sold?  Ans.  25  4,. 

23.  If  $500  gain  $50  in  1  yr.,  in  what  time  will  $900  gain  $60? 
^      24.  Received   an  invoice  of  crockery,   12  per  cent,  of  which  was 
»     broken  ;  at  what  per  cent,  above  cost  must  the  remainder  be  sold,  to 

clear  25  per  cent,  on  the  invoice?  Ans.  42-^^-. 

25.  The  sum  of  two  numbers  is  365,  and  their  difference  is  .0675 ; 
what  are  the  numbers  ? 

26.  If  the  interest  of  $445,621  be  $128.99  for  7  yr.,  what  will  be  the 
interest  of  $650  for  3  yr.  10  mo.  15  da.  ? 

27.  Received  from  Savannah  150  bales  of  cotton,  each  weighing 
540  lb.,  and  invoiced  at  7d.  a  pound  Georgia  currency.  Sold  it  at  an 
advance  of  26  ^,  commission  IJ  ^,  and  remitted  the  proceeds  by 
draft.    What  was  the  face  of  the  draft,  exchange  being  ^  fc  discount? 

Ans.  $12629.28+. 
^  28.  A  man  in  Chicago  haa  5000  francs  due  him  on'account  in  Paris. 
He  can  draw  on  Paris  for  this  amount,  and  negotiate  the  bill  at  19| 
cents  per  franc;  or  he  can  advise  his  correspondent  in  Paris  to  remit 
a  draft  on  the  United  States,  purchased  with  the  sum  due  him,  ex- 
change on  U.  S.  being  at  the  rate  of  5  francs  20  centimes  per  $1. 
What  sum  will  the  man  receive  by  each  method? 

Ans.  By  draft  on  Paris,  $970 ;  by  remittance  from  Paris,  $961.53. 
'\/  29.  What  sum  must  be  invested  in  stocks  bearing  6J  ^,  at  105^, to 
'^^^roduce  an  income  of  $1000?/^^  jl  ...      .      .,     -^ns.  «?16153.84. 

30.  A  person  exchanges  250  shares  of  6  per  cent,  stock,  at  70  fo, 
for  stock  bearing  8  per  cent.,  at  120  ^  ;  what  is  the  difference  in  his 
income?  Ans.  $333. 33J. 

31.  If  f  of  A^s  money  equals  f  of  B's,  and  f  of  B's  equals  'j  of  O's, 
and  the  interest  of  all  their  money  for  4  yr.  8  mo.  at  6  ^  is  $15190, 
how  much  money  has  each  ? 

Ans.  A  has  $18859.44+ ;  B,  $16763.95+ ;  C,  $18626.61. 

32.  A  boy  14  years  old  is  left  an  annuity  of  $250,  which  is  de- 
posited in  a  savings  bank  at  6  ^,  interest  payable  semi-annually; 
how  much  will  he  be  worth  when  of  age?  Ans,  $2104.227. 

33.  If  a  boy  buys  peaches  at  the  rate  of  5  for  2  cents,  and  sells 
them  at  the  rate  of  4  for  3  cents,  how  many  must  he  buy  and  sell  to 
mal^e  a  profit  of  $4.20  ?  /  J^  ^^ 

34.  What  fo  in  advance  of  the  cost  must  a  mercnant  mark  his 
goods,  so  that,  after  allowing  5  j^  of  his  sales  for  bad  debts,  an  ave- 
rage credit  of  6  months,  and  7  ^  of  the  cost  of  the  goods  for  his  ex- 
penses, he  may  make  a  clear  gain  of  12^]  ^  on  the  first  cost  of  the 
goods,  money  bein^  worth  6  ^  ?  Ans.  29.56  •{-  ^ . 


< 


424  MISCELLANEOUS   EXAMPLES. 

35.  Four  men  contracted  to  do  a  certain  job  of  work  for  $8600;  the 
first  employed  28  laborers  20  da.,  10  h.  a  day;  the  second,  25  Laborers 
15  da.,  12  h.  a  day;  the  third,  18  hiborers  25  da.,  11  h.  a  day;  and 
the  fourth,  15  laborers  24  da.,  8  h.  a  day.  How  much  should  each 
contractor  receive  ? 

Ans.  Ist,  $2686;  2d,  $2158.39;  3d,  $2374.24;  4th,  $1381.37. 

36.  If  I  exchange  75  railroad  bonds  of  $500  each,  at  36  %  below 
f/^JCpar,  for  bank  ^ck  at  5  %  premium,  how  many  shares  of  $100  each 

will  I  receive?  Ans.  2281^.    . 

37.  A  trader  has  bought  merchandise  as  follows  :  July  3,  $35.26  ; 
July  4,  $48.65,  on  30  da. ;  Aug.  17,  $6.48 ;  Sept.  12,  $50.  What  is 
due  on  the  account  Oct.  12,  interest  at  9  %  ?  Ans.  142.60. 

38.  A  farmer  sold  34  bu.  of  corn,  and  56  bu.  of  barley  for  $63.10, 
receiving  35  cents  a  bushel  more  for  the  barley  than  for  the  corn ; 
what  was  the  price  of  each  per  bushel? 

39.  A  speculator  purchased  a  quantity  of  flour,  Sept.  1 ;  Oct.  1  its 
value  had  increased  25  %  ;  Nov.  1  its  value  was  30  %  more  than  Oct. 
1;  Deo.  1  he  sold  it  for  15  %  less  than  its  value  Nov.  1,  receiving  in 
payment  a  6  months'  note,  which  he  got  discounted  at  a  bank,  at  7 
%,  receiving  $12950  on  it.     How  much  was  his  profit  on  the  flour? 

Ans.  $3228.51. 

40.  A  flour  merchant  bought  120  bbl.  of  flour  for  $660,  paying 
$5.75  for  first  quality  and  $5  for  second  quality ;  how  many  barrels 
were  first  quality?  ')/  Ans.  80. 

41.  Two  mechanics  work  together ;  for  15  days'  work  of  the  first 
and  8  days'  work  of  the  second  they  receive  $61,  and  for  6  days' 
work  of  the  first  and  10  days'  work  of  the  second  they  receive 
$38  ;  how  much  does  each  man  earn  ?  Ans.  1st,  $63  ;  2d,  $36. 

42.  The  duty,  at  15  %,  on  Rio  cofl'ee,  in  bags  weighing  180  lbs. 
gross,  and  invoiced  at  $.12J  per  pound,  was  $961. 87J,  tare  having  been 
allowed  at  5  %  ;  how  many  bags  were  imported  ?  A7is.  300. 

43.  A  dairyman  took  some  butter  to  market,  for  which  he  received 
$49,  receiving  as  many  cents  a  pound  as  there  were  pounds ;  how 
many  pounds  were  there  ?  Ans.  70  lb. 

44.  A  mechanic  received  $2  a  day  for  his  labor,  and  paid  $4  a  week 
for  his  board ;  at  the  expiration  of  10  weeks  he  had  saved  $72 ;  how 
many  days  did  he  work,  and  how  many  was  he  idle  ? 

45.  To  what  would  $250,  deposited  in  a  savings  bank,  amount  in 
10  yr.,  interest  being  allowed  sem^-|innually  at  6  %  per  annum  ? 

46.  How  much  water  is  there  in^a  mixture  of  100  gal.  of  wine  and 
water,  worth  $1  per  gal.,  if  100  gal.  of  the  wine  cost  $i^^? 

^^  47.  If  a  pipe  3  in.  in  diameter  will  discharge  a  cermm  quantity  of 
water  in  2  h.,  in  what  time  will  3  two-inch  pipes  discharge  3  times 
the   quantity  ?  Ans.  4  h.  30  min. 

48.  Wm.  Jones  &  Co.  become  insolvent  and  owe  $8100.  Their 
assets  amount  to  $4981.50.     What  per  cent,  of  their  indebtedness  can 


MISCELLANEOUS  EXAMPLES.  425 

they  pay,  allowing  the  assignees  2J  ^   on  the  amount  distributed 

for  their  services  ?  Ans,  GO  per  cent. 

49.  Shipped  a  car  load  of  fat  cattle  to  Boston,  and  offered  them  for 

sale  at  25  per  cent.  adAance  on  the  cost ;  but  the  market  being  dull  I 
^  sold  for  14  per  cent,  less  than  my  asking  price,  and  gained  thereby 
/^^  $170.     How  much  did  the  cattle  cost ;  for  how  much  did  they  sell ; 

and  what  was  my  asking  price? 

Ans.  Cost  $22G6.6Gf ;  sold  for  $2436.66f ;  asking  price,  $2833.33 J. 

/V^      50.  What  must  be  the  dimensions  of  a  cubical  cistern  to  hold  2000 

\  51.  A  man  died  leaving  $5000  to  be  divided  between  his  three  sons, 
aged  13,  15,  and  IG  yr.  G  mo.,  respectively,  in  such  a  proportion  that 
the  share  of  each  being  put  at  simple  interest  at  6  %,  should  amount 
to  the  same  sum  when  they  should  arrive  at  the  age  of  21.  How 
much  was  each  one's  share?  <^. 

>^^  Ans.  Youngest,  $1536.76+  ;  second,  $1672.36+  ;  oldest,  $1790.88  +  . 

52.  A  vessel  having  sailed  due  south  and  due  east  on  alternate  days, 
was  found,  after  a  certain  time,  to  be  118.794  miles  south-east  of  the 
place  of  starting  ;  what  distance  had  she  sailed  ?      Ans.  1G8  miles. 

53.  Imported  4  pipes  of  Madeira  wine,  at  $2.15,  a  gallon,  and  paid 
\-  $57.60  freight,  and  a  duty  of  24  per  cent.  I  sold  the  whole  for  $1980 ; 
i'        what  was  my  gain  ^  ? 

54.  If  34J  bu.  of  corn  are  equal  in  value  to  1 7  bu.  wheat,  9  bu.  of 
wheat  to  59 J  bu.  of  oats,  and  6  bu.  of  oats  to  42  lb.  of  flour,  how  many 
bushels  of  corn  will  purchase  5  bbl.  of  flour  ?      y^  Aiis.  42|J|. 

/y       55.    If  stock  bought  at  8  %  discount  will  pay  7  ^  on  tbo  it.^-.  ct„ 
P^  ment,  at  what  rate  should  it  be  bought  to  pay  10  fo  2      '^^ 

5G.  A  merchant  in  New  York  gave  $2000  for  a  bill  of  cxciiaii-f  .a 
X400  to  remit  to  Liverpool ;  what  was  the  rate  in  favor  of  Engfand? 

57.  A,  B,  and  C  start  from  the  same  point,  to  travel  around  a  lake 
84  miles  in  circumference.  A  travels  7  miles,  and  B  21  miles  a  day 
in  the  same  direction,  and  C  14  miles  in  an  opposite  direction.  In 
how  many  days  will  they  all  meet?  Ans.  12. 

58.  The  exact  solar  year  is  greater  than  365  days  by  i^i^^  of  a 
day  ;  find  approximately  how  often  leap  year  should  come,  or  one  day 
be  added  to  the  common  year,  in  order  to  keep  the  calendar  right  ? 

Ans.  Once  in  every  4  yr. ;  7  times  in  every  28  yr. ;  8  times  in  every 
S3  yr. ;  31  times  in  every  128  yr. ;  or  163  times  in  every  673  3^r. 
,  59.  A  gentleman  purchases  a  farm  for  $10000,  which  he  sells  after 
a  certain  number  of  years  for  $14071,  making  on  the  investment  5  fo 
compound  interest.  He  now  invests  his  money  in  a  perpetuity,  which 
is  in  reversion  11  years  from  the  date  of  purchasing  the  farm.  Al- 
lowing 6  fo  compound  interest  for  the  use  of  money,  find  the  annuity 
and  the  length  of  time  he  owns  the  farm. 

y         Ans.  Annuity,  $1065.85  :  owned  the  farm  7  yr. 


426  MISCELLANEOUS  EXAMPLES. 


.  What  will  I  gain  %  by  purchasing  goods  on  6  mo.,  and  selling 
immediately  for  cash  at  cost,  money  being  worth  7  %  ?       4/«^. 


GO. 
them 

Gl.  What  sum  must  a  man  save  annually,  commencing  at  21  years 
of  age,  to  be  worth  $30000  when  he  is  50  years  old,  his  savings  being 
invested  at  5  %  compound  interest?  -f.  Ans.  $481.37. 

62.  Three  persons  are  to  share  $10000 'in  the  ratio  of  3,  4,  and  5, 
but  the  first  dyin^:  it  is  required  to  divide  the  whole  sum  equitably  be- 
tween the  other  two.     What  are  the  shares  of  the  other  two  ? 

Ans.  $4444f,  and  $5555f. 

G3.  If  50  bbl.  of  flour  in  Chicago  are  worth  125  yd.  of  cloth  in  New 
York,  and  80  yd.  of  cloth  in  New  York  are  worth  G  bales  of  cotton  in 
Charleston,  and  13  bales  of  cotton  in  Charleston  are  worth  3 J  hhd. 
of  sugar  in  New  Orleans,  how  many  hhd.  of  sugar  in  New  Orleans 
are  vrorth  1500  bbl.  of  flour  in  Chicago?  Ans.  75  ,^4^. 

G4.  Seven  men  all   start  together  to  travel  the  same  way  round  an 

island  120  miles  in  circumference,  and  continue  to  travel  until  they 

all  come  together  again.     They  travel  5,  G},  7J,  S^,  9J,  10}  and  11} 

miles  a  day  respectively.    In  how  many  days  will  they  all  be  together 

again  ?  Aiis.  1440  da. 

G5.  There  are  two  clocks  which  keep  perfect  time  when  their  pen- 
dulums beat  seconds.  The  first  loses  20  seconds  a  da}^  and  the  second 
gains  15  seconds  a  day.  If  the  two  pendulums  beat  together  when 
both  dials  indicate  precisely  12  o'clock,  what  time  does  each  clock 
show  when  the  pendulums  next  beat  in  concert? 

Ans.  The  first  shows  41  min.  8  sec.  past  12 ;  and  the  second  41 
min.  9  sec.  past  12. 

G6.  If  a  body  put  in  motion  move  J  of  an  inch  the  first  second  of 
time,  1  in.  the  second  sec,  3  in.  the  third,  and  so  continue  to  increase 
in  geometrical  ratio,  how  far  would  it  move  in  30  seconds  ? 

A71S.  541o9u730fMi  mi. 

G7.  If  stock  bought  at  5  fo  premium  will  pay  6  ^  on  the  invest- 
ment, what  fo  will  it  pay  if  bought  at  15  fo  discount  ?    Ans.  Ty^  %* 

G8.  If  G  apples  and  7  peaches  cost  33  cts.,  and  10  apples  and  8 
peaches  cost  44  cts.,  what  is  the  price  of  one  of  each  ? 

A71S,  Apples,  2  cts. ;  peaches,  3  cts. 

GO.  A  gentleman  in  dividing  his  estate  among  his  sons  gave  A  $9 
as  often  as  B  $5,  and  C  $3  as  often  as  B  $7.  .  C's  share  was  5538G2.50 ; 
what  was  the  value  of  the  whole  estate?     '}(  Ans.  $£1,097.50. 

70.  A  farmer  sold  IG  bu.  of  corn  and  20'bu.  of  rye  for  $30,  and  24 
bu.  of  corn  and  10  bu.  of  rye  for  $27.  How  much  per  bushel  did  he 
receive* for  each  ?  ^/  Ans.  Corn,  $.75  ;  rye,  $.90. 

71.  A  drover  sold  some  oxen  at  $28,  cows  at  $17,  and  sheep  at 
$7.50  per  head,  and  received  $749  for  the  lot.  There  were  twice  as 
many  cows  as  oxen,  and  three  times  as  many  sheep  as  cows.  How 
many  were  there  of  each  kind  ?      /Jy    ^'^f  .\.'  i 

72.  For  what  sum  must  a  vessel,  valued  at  $25000,  be  insured,  so 


MISCELLANEOUS  EXAMPLES.  427 

that  in  case  of  its  loss,  the  owners  may  recover  both  the  value  of  the 
vessel  and  the  premium  of  24  f.W2^9p^V?g.  .^v^,    /%>m^^i 

73.  A  boy  hired  to  a  mechani^for  20  weeKs,oav  condition  that  he 
should  receive  $20  and  a  coat.  At  the  end  of  12  weeks  the  boy  quit 
work,  when  it  was  found  that  he  was  entitled  to  $9  and  the  coat ; 
what' was  the  value  of  the  coat?  y!  Ans.  ^l.bO. 

74.  An  irreo-ular  piece  of  land,  containing  540  A.  36  P.,  is  ex- 
chan'o-ed  for  a  "square  piece  containing  the  same  area;  what  is  the 
length  of  one  of  its  sides  ?  If  divided  into  42  equal  squares,  what 
wiUbe  the  length  of  the  side  of  each?       -/ 

75  What  will  be  the  difference  in  the  expense  of  fencing  two  fields 
of  25  acres  each,  one  square,  and  the  other  in  the  form  of  a  rectangle, 
whose  length  is'lwice  its  breadth,  the  fence  costing  $.62i  a  rod? 

/^  Ans.  $9.59+. 

76.  At  what  time  between  5  and  6  o'clock  are  the  hour  and  minute 
hands  of  a  watch  exactly  together  ?  ^  v 

77.  A  general,  forming  his  army  into  a  square,  had  284  men  re- 
maining ;  but  increasing  each  side  by  one  man,  he  wanted  25  men  to 
complete  the  square.     How  many  men  had  he?    /        Ans.  24000. 

78.  Divide  $3618  among  3  persons,  so  that  the  share  of  the  first  to 
^that  of  the  second  shall  be  as  7  to  9,  and  of  the  first  to  the  third  as  3 

to  4.  ♦  Ans.  $1008,  $1296,  $1344. 

^  79.  If  a  lot  of  land,  in  the  form  of  an  oblong  or  rectangle,  contains 

6  A.  3  E.  12  P.,  and  its  length  is  to  its  width  as  21  to  13,  what  are 
its  dimensions ;  and  how  many  rods  of  fence  will  be  required  to  in- 
close it?  Ans.  to  last,  136  rd.  of  fence. 

f\6/  \  80.  Five  persons  are  employed  to  build  a  house.     A,  B,  C,  and  D 

/^can  build  it  in  13  days ;  A,  B,  C,  and  E  in  15  days ;  A,  B,  D,  and  E 

in  12  days;  A,  C,  D,  and  E  in  19  days;  and  B,  C,  D,  and  E  in  14 

days.     In  how  many  days  can  all  together  build  it ;  and  which  cna 

could  do  the  work  alone  in  the  shortest  time  ? 

Ans.  llf*oVj\  da. ;  B  in  shortest  time. 
81.  Divide  $500  among  3  persons,  in  such  a  manner  that  the  share 
of  the  second  may  be  J  greater  than  that  of  the  first,  and  the  share 
of  the  third  J  greater  than  that  of  the  second. 

Ans.  1st,  $105/g;  2d,  $157 jj;  3d,  $236]  J. 

v^  82.  A  and  B  engage  in  trade  ;  A  puts  in  $5000,  and  at  the  end  ofj 

4  mo.  takes  out  a  certain  sum.     B  puts  in  $2500,  and  at  the  end  of  5 

mo.  puts  in  $3000  more.     At  the  end  of  the  year  A's  gain  is  $1066 §, 

and  B's  is  $1333^.     What  sum  did  A  take  out  at  the  end  of  4  mo.  ? 

Ans.  $2400. 

83.  What  sum  of  money,  with  its  semi-annual  dividends  of  5  ^ 
invested  with  it,  will  amount  to  $12750  in  2  yr.  ?  Ans.  ^10489.450-. 

84.  If  a  speculator  invests  $1500  in  flour,  and  pays  5  fo  for  freights, 
2  fo  for  commission,  and  the  flour  sells  at  20  fc  advance  on  cost  price, 
on  a  credit  of  90  days,  and  he  gets  this  paper  discounted  at  bank  at 

7  %,  and  repeats  the  operation  every  15  days,  investing  all  the  pro- 
ceeds each  time,  how  much  will  be  his  whole  gain  in  two  months  ? 


^8  MISCELLANEOUS  EXAMPLES. 

85.  If  a  piece  of  silk  cost  $.80  per  yard,  at  what  price  shall  it  be 
marked,  that  the  merchant  may  sell  it  at  10  %  less  tiian  the  marked 
price,  ahd  still  make  20  %  prorit?  Aas.  $LOGf. 

86.  A  merchant  bought  20  pieces  of  cloth,  each  piece  containing 
25  yd.  at  %\%  per  yard  on  a  credit  of  9  mo. ;  he  sold  the  goods  at 
$4|  per  yard  on  a  credit  of  4  mo.  What  was  his  net  cash  gain, 
money  being  worth  G  ^^  ?  Ans,  $173.85. 

-*  87.  A  owes  B  $1200,  to  be  paid  in  equal  annual  payments  of  $200 
each  ;  but  not  being  able  to  meet  these  payments  at  their  maturities, 
and  having  an  estate  10  years  in  reversion,  he  arranges  with  B  to 
wait  until  he  enters  upon  his  estate,  when  he  is  to  pay  B  the  whole 
amount,  with  8  %  compound  interest.  What  sum  will  B  then  re- 
ceive? Ans,  $1996.074+. 

88.  A  gentleman  who  was  entitled  to  a  perpetuity  of  $3000  a  year, 
provided  in  his  will  that,  after  his  decease,  his  oldest  son  shoiild  receive 
It  for  10  yr.,  then  his  second  son  for  the  next  10  yr.,  and  a  literary 
institution  for  ever  afterward.  What  was  the  value  of  each  bequest 
at  the  time  of  his  decease,  allowing  compound  interest  at  6  ^  ? 

Ans,  To  oldest  son,  $22080.28  ;  to  second  son,  $1^329.51 ;.,  to  insti- 
tution, $15590.23. 

89.  B  has  3  teams  engaged  in  transportation ;  his  horse  team  can 
perform  the  trip  in  5  days,  the  mule  team  in  7  days,  and  the  ox  team 
in  11  days.  Provided  they  start  together,  and  each  team  rests  a  day 
after  each  trip,  how  many  days  will  elapse  before  they  all  rest  the 
same  day?  Ans.  23  days. 

90.  A  man  bought  a  farm  for  $4500,  and  agreed  to  pay  principal 
and  interest  in  4  equal  annual  installments;  how  much  was  the  annual 
payment,  interest  being  6  ^  ?  Ans,  $1298.67  + . 

^  91.  A  bought  a  piece  of  property  of  B,  and  gave  him  his  bond  for 
$6300,  dated  Jan.  1,  1860,  payable  in  6  equal  annual  instalments  of 
$1050,  the  first  to  be  paid  Jan.  1,  1861.  A  took  up  his  bond  Jan. 
1,  1864,  semi-annual  discount  at  the  rate  of  6  %  per  annum  on  the 
two  paynrients  which  fell  due  after  Jan.  1,  1864,  being  deducted; 
what  sum  canceled  the  bond?  J.ns.  $2972.54+. 

92.  A  gentleman  desires  to  set  out  a  rectangular  orchard  of  864  trees, 
so  placed  that  the  number  of  rows  shall  be  to  the  number  of  trees  in  a 
row,  as  3  to  2.  If  the  trees  are  7  yards  apart,  how  much  ground  will 
tlie  orchard  occupy  ?  Ans,  39445  sq.  yd, 

V  93.  S.  C.  Wilder  bought  25  shares  of  bank  stock  at  an  advance  of 

6  %  on  the  par  value  of  $100.  From  the  time  of  purchase  until 
the  end  of  3  yr.  3  mo.  he  received  a  semi-annual  dividend  of  4  ^o, 
when  he  sold  the  stock  at  a  premium  of  11  ^.     Money  being  worth 

7  ^0  compound  interest,  how  much  did  he  gain?  Ans,  $137.31. 


THE  METRIC  SYSTEM 


WEIGHTS    AND    MEASURES.* 


INTRODUCTION. 

Tlie  metric  system  of  weights  and  measures  —  so  called,  because 
the  metre  is  the  unit  from  which  the  other  units  of  the  system  are 
derived — had  its  origin  in  France  during  the  Eevolution,  a  time  when 
all  regard  for  institutions  of  the  past  was  repudiated.  In  the  year 
1790,  the  French  government  resolved  to  introduce  a  new  Fystem ; 
and,  in  order  that  it  might  be  received  with  general  favor,  other 
countries  were  invited  to  join  with  it  in  the  choice  of  new  units. 
In  response  to  this  invitation,  a  large  number  of  scientific  men,  com- 
missioned by  various  countries,  met  in  Paris,  in  consultation  with  the 
principal  men  of  France.  In  the  year  1791,  a  commission,  nomi- 
nated by  the  Academy  of  Sciences,  was  appointed  by  the  Government 
to  prepare  the  new  system.  The  first  work  of  the  commission  was  to 
select  a  standard  of  lengths  from  which  the  system  of  units  adopted 
might  at  any  time  be  restored  if  from  any  cause  the  original  unit 
should  be  lost.  A  quadrant  of  the  earth's  meridian  was  chosen  as 
the  standard,  and  the  ten-millionth  part  of  it  taken  as  iha  unit  cf 
lengths,  which  was  called  a  metre.  In  1795,  this  standard  and  a 
provisional  metre  whose  length  was  determined  from  measurements 

*  31.  McYiCAR,  A.M.,  Principal  of  the  State  Normal  and  Training  School  at 
Brockport,  N.Y.,  a  most  thorough  and  critical  scholar  as  well  as  teacher,  prepared 
this  article,  which  contains  many  practical  improvements  in  Notation,  Nomencla- 
ture, and  Applications,  not  before  presented  to  the  public.  . 

Entered,  according  to  Act  of  Congress,  in  the  year  1867,  by  D.  W.  Fisn,  A.  M.,  in  the  Clerk'tt 
Office  of  the  District  Ck>urt  of  the  United  States  for  the  Southern  District  of  New  York. 

(429) 


430  THE   METRIC   SYSTEM. 

of  the  earth's  meridian,  "which  had  already  been  made,  was  adopted 
by  the  government. 

In  the  meantune,  two  eminent  astronomers,  Mechain  and  Delambre, 
were  engaged  in  determining  the  exact  length  of  the  arc  of  the  meri- 
dian between  Dunkirk  in  the  north  of  France,  and  Barcelona  in 
Spain.  At  a  later  period,  Liot  and  Arago  measured  the  prolonga- 
tion of  the  same  meridian  as  far  as  the  island  of  Formentara.  From 
these  measurements,  together  with  one  formerly  made  in  Peru,  they 
deduced,  as  they  supposed,  the  exact  distance  from  the  equator  to 
the  pole,  which  differed  slightly  from  the  standard  assumed  in  1795. 
In  1790,  a  law  was  passed  changing  the  length  of  the  metre  adopted 
in  1795  so  as  to  conform  with  this  diSerence.  The  metre  thus  de- 
termined was  marked  by  two  very  fine  parallel  lines  drawn  on  a  pla- 
tinum bar,  and  deposited  for  preservation  in  the  national  archives. 

While  a  part  of  the  commission  were  engaged  in  establishing  the 
exact  length  of  the  metre,  other  members  pursued  a  course  of  inves- 
tigation for  the  purpose  of  determining  a  unit  of  weights,  which  would 
sustain  an  invariable  relation  to  the  unit  of  lengths.  As  the  result 
of  their  investigations,  the  weight  of  a  cube  of  pure  water  whose  edge 
was  one-hundredth  part  of  a  metre  was  the  unit  chosen.  The  water 
was  weighed  in  a  vacuum,  at  a  temperature  of  4°  C,  or  39.2°  F., 
which  was  supposed  to  be  the  temperature  of  greatest  density.  This 
weight  was  called  a  gramme ^  and  a  piece  of  platinum  weighing  one 
thousand  grammes  was  deposited  as  the  standard  of  weights  in  th^ 
national  archives. 

Had  the  work  of  the  commission  ended  in  determining  these 
standards  of  lengths  and  weights,  their  labor  would  have  been  futile. 
For,  while  the  conception  of  basing  their  system  upon  an  absolute 
standard  in  nature  was  good,  the  execution  proved  a  failure.  Later 
in vesti stations  have  shown  that  the  metre  is  less  than  the  ten-millionth 
part  of  the  earth's  meridian ;  consequently  the  metric  system  of 
weights  and  measures  is  referable  not  to  an  invariable  standard  in 
nature,  but  to  the  platinum  metre  deposited  in  the  national  archives 
of  France.  The  great  benefits  which  result  from  the  labors  of  tlie 
commission  arise  from  the  adoption  of  the  decimal  scale  cf  units,  and 
a  simple  yet  general  and  expressive  nomenclature.     The  amount  of 


THE   METRIC   SYSTEM.  431 

time  and  money  nsed  in  carrying  on  exchanges  between  different  coun- 
tries, Yrliich  would  be  saved  by  the  universal  adoption  of  this  system, 
is  incalculable.  The  system  was  declared  obligatory  throughout  the 
whole  of  France  after  Xov.  2,  1801 ;  but,  owing  to  the  prejudices 
of  the  people  in  favor  of  established  customs,  and  the  confusion  con- 
eequent  upon  the  use  cf  the  new  measures,  the  Government,  in  1812, 
adopted  a  compromise,  in  the  systeme  usuile,  whose  principal  units 
were  the  new  ones,  while  the  divisions  and  names  were  nearly  those 
formerly  in  use,  ascending  commonly  in  the  ratios  of  two,  three,  four, 
eight,  or  twelve.  In  1837,  the  government  abolished  this  system, 
and  enacted  a  law  attaching  a  penalty  to  the  use  of  any  other  than 
the  metric  system  after  Jan.  1,  1841.  Since  that  time,  the  system 
has  been  adopted  by  Spain,  Belgium,  and  Portugal,  to  the  exclusion 
of  other  weights  and  measures.  In  Holland,  other  weights  are  used 
only  in  compounding  medicines.  In  18G4,  the  system  was  legalized 
in  Great  Britain ;  and  its  use,  either  as  a  whole  or  in  some  of  its  parts, 
has  been  authorized  in  Greece,  Italy,  Norway,  Sweden,  Mexico, 
Guatemala,  Venezuela,  Ecuador,  United  States  of  Columbia,  Brazil 
Chili,  San  Salvador,  and  Argentine  Republic.  In  1866,  Congress 
authorized  the  metric  system  in  the  United  States  by  passing  the  fol- 
lowing; bills  and  resolution  ;  — ■ 

An  Act  to  atttiiorize  the  tjse  of  tiie  Metric  System  of  Weights 
AND  Measures. 

Be  it  enacted  hy  the  Senate  and  House  of  Representatives  of  the  United  Stauts 
of  America  in  Congress  assembled,  That,  from  and  after  the  passage  of  this 
Act,  it  shall  be  lawful  throughout  the  United  States  of  America  to  employ 
the  Weights  and  Measures  of  the  Metric  System  ;  and  no  contract  or  dealing, 
or  pleading  in  any  court,  shall  be  deemed  invalid,  or  liable  to  objection,  be- 
cause the  weights  or  measures  expressed  or  referred  to  therein  arc  weights  or 
measures  of  the  Metric  System. 

Section  2.  And  he  it  farther  enacted,  That  the  tables  in  the  schedule 
hereto  annexed  shall  be  recognized  in  the  construction  of  contracts,  and  in 
all  legal  proceedings,  as  establishmg,  in  terms  of  the  weights  and  measures 
now  in  use  in  the  United  States,  the  equivalents  of  the  weights  and  measures 
expressed  therein  in  terms  of  the  Metric  System ;  and  said  tables  may  be 
lawfully  used  for  computing,  determining,  and  expressing  in  customary 
weights  and  measures,  the  weights  and  measures  of  the  Metric  System. 


432 


THE   METRIC   SYSTEM. 


A  Bill  to  authorize  the  Use  in  Post  Offices  of  the  Weights 
OF  THE  Denomination  of  Grammes. 

Be  it  enacted  by  the  Senate  and  House  of  Representatives  of  the  United  States 
of  America  in  Congress  assembled^  That  the  Postmaster  General  be,  and  he  is 
hereby,  authorized  and  directed  to  furnish  to  the  post-offices  exchanging 
mails  with  foreign  countries,  and  to  such  other  offices  as  he  shall  think  expe- 
dient, postal  balances  denominated  in  grammes  of  the  metric  system  ;  and, 
until  otherwise  provided  by  law,  one-half  ounce  avoirdupois  shall  be  deemed 
and  taken  for  postal  purposes  as  the  equivalent  of  fifteen  grammes  of  the 
metric  weights,  and  so  adopted  in  progression;  and  the  rates  of  postage  shall 
be  applied  accordingly. 

Joint  Resolution  to  enable  the   Secretary  of  the   Treasury 
TO   furnish  to  each  State  one  set  of  the  Standard  Weights 

AND   JVIeASURES    of   THE    MeTRIO    SySTEM. 

Be  it  resolved  by  the  Senate  and  House  of  Representatives  of  the  United  States 
of  America  in  Congress  ossejnbled,  That  the  Secretary  of  the  Treasury  be,  and 
he  is  hereby,  authorized  and  directed  to  furnish  to  each  State,  to  be  delivered 
to  the  governor  thereof,  one  set  of  the  standard  weights  and  measures  of  the 
metric  system,  for  the  use  of  the  States  respectively. 

TABLES  AUTHORIZED  BY  CONGRESS. 
MEASURES  OF  LENGTHS. 


Metric  Denominations  and  Values. 

Equivalents  in  Denominations  in  use. 

Myriametre, . . . 

10,000  metres, 

6.2137  miles. 

Kilometre,  .... 

1,000  metres, 

0.62 137  miles,  or  3280  feet,  10  inches. 

Hectometre , . . . 

100  metres, 

328  feet  and  1  inch. 

Decametre,  . . . 

10  metres, .. 

393.7  inches. 

Metre, 

1  metre, 

39.37  inches. 

Decimetre, .... 

^Q-  of  a  metre,  . . 

3.937  inches. 

Centimetre,  . . . 

■ji^  of  a  metre,  . . 

0.3937  inch. 

Millimetre, 

T  (tW  ^^  ^  ^^^^^'  •  • 

0.0394  inch. 

MEASURES  OF  SURFACES. 


Metric  Denominations  and  Values. 

Equivalents  in  Denominations  in  use. 

Hectare, 

Are, 

Centiare, 

10,000  square  metres, 

100  square  metres, 

1  square  metre, 

2.471  acres. 

119.6  square  yards. 

1550  square  inches. 

THE   METRIC   SYSTEM. 


433 


MEASURES  OF  CAPACITY. 


Metric  Denominations  and  Values. 

Equivalents  in  Denominations  in  use. 

Names. 

No.  of 
litres. 

Cubic  Measure. 

Dry  Measure. 

Liquid  or  wine 
measure. 

Kilolitre,  or  stere. 

Hectolitre,  

Decalitre, 

Litre, 

lOOOjl  cubic  metre, 

100|  jl^  of  a  cubic  metre, . . . 

10  10  cubic  decimetres,. . . 

1  1  cubic  decimetre 

1.308  cubic  yd. 
2  bu.  3.35  pk... 
9.08  quarts,.... 
0.908  quart,  . . . 
6.1022  cubic  in. 
0.6102  cubic  in. 
0.061  cubic  in. . 

264.17  gallon. 
26.417  gallon. 
2.6417  gallon. 
1.0567  quart. 
0.845  gill. 
0.338  fluid  oz. 
0.27  fluid  dr. 

Decilitre, 

Centilitre, 

Millilitre, 

-j\j-  of  a  cubic  decimetre, 
10  cubic  centimetres, . . 
1  cubic  centimetre, 

WEIGHTS. 


Metric  Denominations  and  Values. 

Equivalents  in  De- 
nominations in  use. 

Names. 

Number  of 
grammes. 

Weight  of  what  quantity  of  water 
at  maximum  density. 

Avoirdupois  weight. 

Millier,  or  toimeau, . 
Quintal, 

1,000,000 

100,000 

10,000 

1,000 

100 

10 

1 
iV 

iJo 

TOTT^ 

1  cubic  metre, 

2204.6  pounds. 
220.46  pounds. 
22.046  pounds. 
2.2046  pounds. 
8.5274  ounces. 
0.3527  ounce. 
15.432  grains. 
0.5432  gi-ain. 
0.1543  grain. 
0.0154  gi-ain. 

1  hectolitre, 

Myriagramme, 

Kilogramme,  or  kilo. 

Hectogramme, 

Decagramme, 

Gramme, 

10  litres, 

1  litre,  

1  decilitre, 

10  cubic  centimetres, 

1  cubic  centimetre, 

1-10  of  a  cubic  centimetre, 

10  cubic  millimetres, 

1  cubic  millimetre, 

Decigramme, 

Centigramme, 

Milligramme, 

Note.  —  The  spelling  in  the  above  tables  is  not  the  same  as  in 
the  tables  in  the  schedule  annexed  to  the  report  of  the  committee  of 
the  House  of  llepresentatives  on  weights  and  measures.  The  change 
is  not  made  to  indicate  any  preference  for  any  standard  upon  this 
subject ;  but  to  carry  out  what  the  author  believes  to  be  an  essential 
condition  to  the  utility  and  success  of  the  system. 

As  remarked  by  a  distinguished  senator  when  the  tables  were 
adopted  by  Congress,  '"'-The  names  are  cosmopolitan  ;^''  and  to  re- 
tain this  character  fully,  the  spelling  must  also  he  cosmopolitan. 

The  French  introduced  the  nomenclature  and  spelling ;  and,  so 
long  as  the  names  remain  unchanged,  the  spelling  should  be  retained. 


434 


THE   METRIC   SYSTEM. 


NOMENCLATURE  AND   TABLES. 

t|  Tbcre  are  eight  kinds  of  quantities  for  wbich  tables  are  usnally 
constructed;  viz.,  Lengths,  Surfaces,  Volumes  or  Solids,  Capacities, 
Weights,  Values,  Times,  and  Angles  or  Arcs.  The  table  for  Times 
is  the  same  in  the  metric  as  in  the  ordinary  system.  The  table  for 
Angles  is  constructed  upon  a  centesimal  scale.  The  tables  for  the 
other  six  kinds  of  quantities  are  constructed  upon  a  decimal  scale. 
In  each  of  the  tables  for  Lengths,  Surfaces,  Volumes,  Capacities,  and 
Weights,  there  are  eight  denominations  of  units,  —  one  principal  and 
seven  derivative.  The  principal  units  are  the  metre,  which  is  the 
base  of  the  system,  and  those  derived  directly  from  it.  The  two 
following  tabular  views  present  the  facts  regarding  the  principal  and 
derivative  units,  which  should  be  fixed  in  the  memory. 

"  1.  Principal  unit  of  Lengths. 

2.  The- base  of  the  metric  system,  and  nearly 
one  ten-millionth  part  of  a  quadrant  of 
the  earth's  meridian. 

3.  Equivalent,  39.3708  inches. 

1.  Principal  unit  of  surfaces. 

2.  A  square  whose  side  is  ten  metres. 

3.  Equivalent,  119.6  square  yards. 

1.  Principal  unit  of  volumes  or  solids. 

2.  A  cube  whose  edge  is  one  metre. 

3.  Equivalent,  1.308  cubic  yards. 

1.  Principal  unit  of  capacities. 

2.  A  vessel  whose  volume  is  equal  to  a  cube 
whose  edge  is  one-tenth  of  a  metre. 

3.  Equivalent,  .908  quart  dry  measure,  or 
1.0567  quarts  wine  measure. 

f  1.  Principal  unit  of  weights. 

2.  The  weight  of  a  cube  of  pure  water  whose 
edge  is  .01  of  a  metre. 

3.  The  water  must  be  weighed  in  a  vacuum 
4°  C,  or  39.2°  F. 

^  4.  Equivalent,  15.432  grains. 


«2 

H 

o 

P5 
p-l 


I.  Metre,  . 


II.  Are, 


in.  Stere, 


IV.  Litre,  . .  ^ 


L    V.  Gramme, 


THE  METRIC  SYSTEM. 


435 


ft 

o 


o 
o 


IS? 


fa  >5 
c  *^ 
as  ^^ 
|2l 

O  g  I 
M  o 


2  o. 


1.  Three  orders  of  smaller  units,  or  submultiples  of  each 
kind,  are  formed  by  dividing  each  of  the  principal  units 
into  tenths,  hundredths,  and  thousandths. 

2.  Four  orders  of  larger  units,  or  multiples  of  each  kind, 
are  formed  by  considering  as  a  unit  ten  times,  one 
hundred  times,  one  thousand  times,  and  ten  thousand 
times,  each  of  the  principal  units. 

"  The  names  of  derivative  units  are  formed  by 
attaching  a  prefix  to  the  name  of  the  princi- 
pal unit  from  which  they  are  derived,  which 
indicates  their  relation  to  the  principal  unit. 

1.  Millesimus,  one  thousandth,  contracted 
Milli.  Example,  Millilitre  =  j^ViJ  ^^  ^  ^i^^^ '? 
8  millilitres  =  j^%jj  of  a  litre. 

2.  Centesimus,  one  hundredth,  contracted 
centi.  Ux.,  Sentiare  =  jo (j  ^^  ^^  ^^^i  ^ 
centiares  =  yj^j  of  an  are. 

3.  Decimus,  tenth,  contracted  deci.  jSJx.,  De- 
cimetre =  ^  metre  ;  3  decimetres  =  j\ 
metre. 

1.  Deca,  ten.  Example,  Decametre  =  10 
metres  ;  5  decametres  =  50  metres. 

2.  Hecaton,  one  hundred,  contracted  hecto. 
Ex,,  Hectolitre  =  100  litres ;  7  hectolitres 
=  700  litres. 

3.  Kilioi,  one  thousand,  contracted  kilo.  Ex. 
Kilogramme  =  1000  grammes. 

4.  Myria,  ten  thousand.  Ex.,  Myriastere  = 
10,000  steres;  3  myriasteres  =30,000  stores. 

5.  The  a  in  deca  and  myra,  and  the  o  in  hecto 
and  kilo,  are  dropped  when  prefixed  to  are. 

^  The  tables  being  constructed  upon  a  decimal  scale,  ten 
units  of  a  lower  order  make  one  of  the  next  higher, 
thus:  10  millimetres  =  1  centimetre;  10  centimetres 
=  1  decimetre ;  10  decimetres  =  1  metre ;  10  me- 
tres ==  1  decametre,  &c. 


to    CO 


S  9 
•9  S 


o  fl 


U     03 

3.  O  2  -( 


g  s 


436 


THE   METRIC  SYSTEM. 


The  facts  in  tbe  preceding  views  being  mastered,  tlie  tables  can  be 
constructed  bj  the  pupil  at  sight.  For  example  :  The  names  of  the 
derivative  units  are  formed  by  attaching  the  seven  prefixes,  in  their 
order,  to  the  principal  units  of  the  tables.  The  order  of  progression 
being  ten,  the  table  of  capacities  will  be  written  thus  :  — • 

10  Millilitres   =  1  Centilitre.  10  Litres  =  1  Decalitre. 

10  Centilitres  =  1  Decilitre.  10  Decalitres    =  1  Hectolitre. 

10  Decilitres    =  1  Litre.  10  Hectolitres  =  1  Kilolitre. 

10  Kilolitres  =  1  Myrialitre. 

All  the  tables  peculiar  to  the  Metric  System  are  presented  together 
in  a  convenient  form  in  the  two  following  tables  :  — 

TABLE  OF  SUBMULTIPLES  AND  PRINCIPAL  UNITS. 


Names  of  Units. 

T*  ROX  TINT"  T  A  TT  OV 

Symbols. 

PREFIX. 

BASE. 

JL  £h\J^^  \J  XI  Vy  JL^V  X  X\^^^  f 

p  Metre 

Miir-e-mee'-ter 

3M 

10  Milli- 

Are 

Miir-e-are 

A 

Equal 

Stere 

MilI'-©-ster 

sS 

1  Centi- 

Litre 

Miir-e-li'-ter 

gL 

.  Gramme 

Mill^-e-gram 

8^ 

r  Metre 

Sent^-e-mee'-ter 

2^ 

10  Centi- 

Are 

Sent'-e-are 

2^ 

Equal 

Stere 

Sent'-e-ster 

sS 

1  Deci- 

Litre 

Sent'-e-li'-ter 

^ 

.  Gramme 

Sent^-e-gram 

2<5 

-  Metre 

Des'-e-mee'-tcr 

jM 

le  Deci- 

Are 

Des'-e-are 

1^ 

Equal 

Stere 

Des^-e-ster 

iS 

1  Principal  Unit. 

Litre 

Des^-e-li'-ter 

iL 

-  Gramme 

Des'-e-gram 

6 

-  Metre 

Mee'ter 

M 

10  Principal  Units 

Are 

Are 

A 

Equal 

Stere 

Ster 

S 

1  Deca- 

Litre 

Li'-ter 

L 

-  Gramme 

Gram 

G 

THE   METRIC   SYSTEM. 
TABLE    OF   MULTIPLES. 


437 


Namks  of  Units. 

T*ROXrTVrTATTOV 

PREFIX. 

BASE. 

X    XW^i  O  ■'■I  v/X^x  X  ±\J^  . 

r  Metre 

Dek^-a-mee-ter 

'M 

10  Deca- 

Are 

Dek'-are 

>A 

Equal    -< 

Store 

Dek'-a-ster 

's 

1  Hecto- 

Litre 

Dek'-a-li'-ter 

'l 

.  Gramme 

Dek'-a-gram 

'G 

-  Metre 

Hec^-to-mee-ter 

'm 

10  Hecto. 

Are 

Hec'-tare 

'a 

Equal   - 

Stere 

Hec'-to-ster 

's 

I  Kilo- 

Litre 

Hec'-to-li'-ter 

'l 

*  Gramme 

Hec'-to-gram 

'Gr 

r  Metre 

Kiir-o-mee-ter 

\l 

10  Kilo- 

Are 

Kiir-are 

\ 

Equal    - 

Stere 

KilK-o-ster 

■  's 

1  Myria- 

Litre 

Kill'-o-li^-ter 

'l 

.  Gramme 

Kill'-o-gram 

=G 

-  Metre 

Mir'-e-a-mee-ter 

*M 

Are 

Mir^-e-are 

*A 

Myria-  - 

Stere 

Mir'-e-a-ster 

*S 

Litre 

Mir^-e-a-li'-ter 

*h 

^  Gramme 

Mir'-e-a-gram 

'G 

ABBREVIATED    NOMENCLATUUE. 

To  secure  the  fullest  advantage  to  business  men  by  the  universal 
adoption  of  the  new  system  of  weights  and  measures,  it  is  necessary 
that  the  names  used  should  be  short  and  easy  to  write  and  pronounce, 
that  they  should  express  clearly  the  relation  of  the  different  denomi- 
nations of  the  same  table  to  each  other,  and  that  they  should  be 
identical  in  all  languages. 

The  last  two  of  these  requirements  would  be  secured  by  the  uni- 
versal use  of  the  nomenclature  adopted  by  the  French.  It  is  cosmo- 
politan in  its  character :  it  belongs  to  their  language  no  more  than  to 
any  other.*-  The  former,  however,  is  not  secured.  It  is  evident  to 
all,  that,  for  business  purposes,  the  long  names  of  the  metric  system 
are  inconvenient,  and  that  to  shorten  them  would  prove  a  great 


438  THE  METRIC   SYSTEM, 

advantage.  Efforts  have  been  made  to  introduce  short  names; 
but  these  efforts  have  invariably  sacrificed  their  universal  and  expres- 
sive character,  which  is  of  more  importance  to  the  business  world 
than  their  shortness. 

The  only  true  course  which  seems  to  be  open,  is  to  abbreviate  the 
names  already  introduced,  in  such  a  way  as  to  retain  their  peculiar 
charapteristics. 

To  secure  this,  the  following  plan  of  abbreviation  is  suggested  :  — 

First.  Let  the  prefixes  be  abbreviated  thus  :  Myr,  kil,  hect,  dec, 
des,  cent,  mil. 

Second.  Let  the  initial  letter  of  the  names  of  the  five  principal 
units  be  used,  instead  of  the  names  themselves,  thus  :  For  metre,  use 
a  capital  M ;  for  are,  use  a  capital  A ;  for  store,  a  capital  S  ;  for 
litre,  a  capital  L ;  and,  for  gramme,  a  capital  Gr. 

Third.  For  the  names  of  multiples  and  sub-multiples,  attach  to 
these  initial  capital  letters  the  abbreviated  prefixes,  thus  ;  Kil  M,  pro- 
nounced kill-em'^ ;  Kil  S,  pronounced  kill-ess',  &c. 

By  this  method  of  abbreviation,  the  elements  of  the  original  terms 
are  retained  in  such  a  form  that  each  part  is  clearly  indicated.  The 
capital  letter  used  after  the  prefix  will  always  point  to  the  base-word 
of  which  it  is  the  initial,  although  the  pronunciation  is  changed. 

TABLES  WITH  ABBREVIATED  NOMENCLATURE. 
MEASUEES    OF   LENGTHS. 


Written. 

Pronounced. 

10  Mil  M, 

Mill-em', 

make 

1  Cent  M. 

10  Cent  M, 

Cent-em', 

1  Des  M. 

10  Des  M, 

Des-em' 

1  M. 

10  M, 

Em 

1  Dec  M. 

10  Dee  M, 

Dek-em', 

1  Hect  M. 

10  Hect  M, 

Hect-em', 

1  Kil  M. 

10  Kil  M, 

Kill-em', 

1  Myr  M.* 

Jlyr  M, 

Mir-em'. 

THE   METRIC    SYSTEM. 


439 


MEASURES  OF  SURFACES. 


Written. 

Pronounced. 

10  Mil  A, 

Mill-ii',          make      1  Cent  A 

10  Cent  A, 

Cent-a', 

1  Des  A. 

10  Des  A, 

Des-a', 

1  A. 

10  A, 

A, 

IDccA. 

10  Dec  A, 

Dek-a', 

1  Hect  A 

10  Hect  A, 

Hect-a', 

1  Kil  A. 

10  Kil  A, 

KiU-a', 

1  Myr  A. 

Myr  A, 

Mir-a'. 

MEASURES  OF  VOLUMES,  OR  SOLIDS. 


Written. 

Pronounced. 

10  Mil  S, 

Mill-ess',       make      1  Cent  S. 

10  Cent  S, 

Cent-ess', 

1  Des  S. 

10  Des  S, 

Des-ess', 

IS. 

10  S, 

Ess, 

1  Dec  S. 

10  Dec  S, 

Dek-ess', 

1  Hect  S. 

10  Hect  S, 

Hect-ess', 

1  Kil  S. 

10  Kil  S, 

Kill-ess', 

1  Myr  S. 

Myr  S, 

Mir-ess'. 

MEASURES  OF  CAPACITY. 


Written. 

Pronounced. 

10  Mil  L, 

Mill-eir,       make      1  Cent  L. 

10  Cent  L, 

Cent-eir, 

1  Des  L. 

10  Des  L, 

Dess-ell' 

IL. 

10  L, 

Ell, 

1  Dec  L. 

10  Dec  L, 

Dek-ell', 

1  Hect  L. 

10  Hect  L, 

Hect-eir, 

1  Kil  L. 

10  Kil  L, 

Kill-eir, 

1  Myr  L. 

Myr  L, 

Mir-ell'. 

440  THE   METRIC   SYSTEM. 


MEASURES    OF   WEIGHTS. 


Written. 

Pronounced. 

10  Mil  G, 

Mill-gee', 

make 

1  Cent  G. 

10  Cent  G, 

Cent-gee', 

1  Des  G. 

10  Des  G, 

Des-gee', 

IG. 

10  G, 

Gee, 

1  Dec  G. 

10  Dec  G, 

Dek-gee', 

1  Hect  G. 

10  Hect  G, 

Hect-gee', 

1  Kil  G. 

lOKilG, 

Kill-gee', 

1  Myr  G. 

MyrG, 

Mir-gee'. 

NOTATION   AND    NUMERATION. 

In  the  practical  application  of  the  metric  system,  it  is  not  always 
convenient  to  use  the  principal  units  as  the  unit  of  number.  For 
example  :  Should  the  gramme,  the  principal  unit  of  weight,  be  used 
as  the  unit  of  number,  in  the  grocery  or  any  similar  business,  small 
quantities  would  be  expressed  by  inconveniently  large  numbers. 
Example  :  386  lbs.  are  expressed  by  175,000  grammes.  To  avoid 
this  inconvenience,  the  higher  denominations  aie  used  as  the  unit  of 
number  when  large  quantities  are  measured. 

No  general  system  of  notation  is  yet  agreed  upon.  The  same 
quantity  is  written  in  various  ways  by  different  authors.  Example  ; 
42  metres,  8  decimetres,  and  5  centimetres,  are  written 

m     cm 

42.85  M.        42?  85.        42.85.        M  42  85.    &c. 

Inasmuch  as  the  principal  units  of  measure  are  not  always  used  as 
the  unit  of  number,  it  is  important  that  a  system  of  notation  be  adopt- 
ed, which  will  apply  equally  well  to  both  principal  and  derivative 
units. 

It  is  believed  that  the  system  given  below,  while  simple  and  con- 
venient, expresses  clearly  the  relation  of  the  unit  of  number  to  the 
principal  unit  of  measure ;  and,  hence,  has  an  advantage  over  any 
contractions  of  the  names  of  the  derivative  units  or  arbitrary  signs 
which  might  be  adopted. 


THE   METRIC   SYSTEM.  441 

GENERAL   PRINCIPLES    OF   NOTATION. 

I.  The  scale  in  the  metric  system  being  decimal,  the  consecutive 
denominations  are  expressed  by  the  consecutive  orders  of  units  in  a 
number.  Thus,  78G42.358  metres  is  an  expression  for  7  myria- 
metres,  8  kilometres,  6  hectometres,  4  decametres,  2  metres,  3  deci- 
metres, 5  centimetres,  8  millimetres. 

II.  Whichever  one  of  the  eight  denominations  of  units  of  measure 
is  used  as  the  unit  of  a  number,  the  higher  denominations  are  ex- 
pressed as  tens,  hundreds,  and  so  on ;  and  the  lower  as  tenths,  hun- 
dredths, and  so  on.  Example  :  784.56  decametres.  Here  the  unit 
of  the  number  is  a  decametre ;  consequently  the  tens  and  hundreds 
are,  respectively,  hectometres  and  kilometres,  and  the  tenths  and 
hundredths  are  metres  and  decimetres. 

From  these  principles  and  illustrations,  we  derive  the  following  rule 
for  notation  :  — 

Rule.  Write  the  consecutive  denominations  in  their  order,  com- 
mencing  with  the  higher,  and  placing  a  cipher  wherever  a  denomi- 
nation is  omitted,  and  the  decimal  point  after  the  denomination 
which  is  the  unit  of  the  number. 

KULES    FOU   INDICxVTING   THE   DENOMINATION. 

KuLE  I.  When  a  principal  unit  of  measure  is  the  unit  of  Clum- 
ber, place  the  initial  letter  of  the  unit  used  before  the  number,  thus : 
M  342.5.  Read,  three  hundred  and  forty-two  and  five-tenths 
metres  ;  or^  3  hectometres,  4  decametres,  2  metres,  5  decimetres, 

EXAMPLES    FOR    PRACTICE. 

Write  the  numbers  which  represent  the  following  quantities,  con- 
sidering the  principal  unit  of  measure  the  unit  of  number. 

1.  Seven  myriametres,  4  hectometres,  three  decametres,  and  eight 
centimetres.  Ans.  M  70430.08. 

2.  Thirty-four  kilometres  and  forty-three  millimetres. 

Ans,  M  34000.043. 

3.  Eighty-seven  hectogrammes  and  fifty-nine  centigrammes. 

Ans:  G  8700.59. 


442  THE   METRIC   SYSTEM. 

4.  Thirty-two  myriagrammes,  forty-eigbt  decagrammes,  five  milli- 
grammes. Ans.  a  320480.005. 

5.  Three  hundred  and  two  kilares,  eight  hundred  and  seven  cen- 
tiares.  Ans.  G  302008.07. 

6.  Four  myrialitres,  sixty-two  decalitres,  live  millilitres. 

Ans,  L  40620,005. 

7.  Four  hundred  and  thirty-three  kilosteres,  nine  hundred  and 
eighty  four  hectosteres,  seven  thousand  two  hundred  and  three  centi- 
steres.  A?is.  S  53147203. 

EuLE  II.  When  a  multiple  of  a  principal  unit  of  measure  is  the 
unit  of  number ;  —  First,  Place  before  the  number  the  initial  letter 
of  the  principal  unit  from  ivhich  the  multiple  is  derived.  Second, 
Indicate  the  order  of  multiple  used  by  a  small  figure  placed  to  the 
left  and  above  the  letter  prefixed  to  the  number.  (See  symbols  in 
table  of  multiples.) 

Example.    42.5  kilometres,  is  written  ^M42.5. 

The  M  before  the  number  indicates  that  the  metre  is  the  unit  of 
measure  from  which  the  unit  of  the  number  is  derived.  The  small 
3  indicates  that  the  third  order  of  multiple,  or  kilometre,  is  the  unit 
of  number. 

EXAMPLES   FOR   PRACTICE. 

Write  the  numbers  which  represent  the  following  quantities,  con- 
sidering the  denomination  named  as  the  unit  of  number  :  — 

Unit  of  Number,  Kilogramme, 

1.  43  myriagrammes,  7  decagrammes,  5  grammes. 

Ans.  ^G  430.075. 

2.  8  kilogrammes  and  3  centigrammes.        Ans.  ^G  8.00003. 

3.  736  hectogrammes,  243  centigrammes,  and  4  milligrammes. 

Ans.  ^G  73.602434. 

4.  2009  hectogrammes  and  3  centigrammes. 

Ans.  ^G  200.90003. 

Ufiit  of  Number ,  Decalitre. 

5.  254  litres  and  43  milUlitres.  Ans.  ^L  25.4043. 


THE   METPJC   SYSTEM.  443 

6.  364  mjrialitres,  47  litres,  384  millilitres. 

^;is.  1L3G4004.7384. 

7.  243  decalitres,  47  centilitres.  Ans,  ^L  243.047. 

Unit  of  Number,  Second  Order  of  3Iultiples. 

8.  23  myriametres,  72  millimetres.         Ans.  ^31  2300.00072. 

9.  -4000007  steres  and  2  millisteres.     Ans,  ^S  40000.07002. 

10.  3  kilares  and  43  centiares.  Ans,  ^A  30.0042. 

Unit  of  Numher,  Myriametre. 

11.  3  hectometres  and  2  centimetres.  Ans.  "^M  .030002. 

12.  5  millimetres.  Ans.  ^xM  .0000005. 

13.  3  decametres  and  2  centimetres.  Ans.  ^M  .003002. 

Rule  III.  When  a  submultiple  of  a  principal  unit  of  measure 
is  the  unit  of  number ;  —  First,  Place  before  the  number  the  initial 
letter  of  the  principal  unit  from  which  the  submultiple  is  derived. 
Second,  Indicate  the  order  of  submultiple  used  by  a  small  figure 
placed  to  the  left  and  below  the  letter  prefixed  to  the  number.  (See 
symbols  in  table  of  submultiples.) 

EXAMPLES   FOR   PRACTICE. 

Write  the  numbers  which  represent  the  following  quantities,  con- 
sidering the  denomination  named  as  the  unit  of  number. 

Unit  of  Number,  Millimetre. 

1.  32  decametres  and  2  decimetres.  Ans.  gM  320200. 

2.  7002  hectometres.  Ans.  .M  700200000. 

3.  7  myriametres  and  5  metres.  Ans.  gM  70005000. 

4.  3  kilometres  and  2  decametres.  Ans.  3M  3020000. 

Unit  of  Number,  Second  Order  of  Submultiples. 

5.  5  kilogrammes  and  9  grammes.  Ans.  gCr  500900. 

6.  302  myriasteres,  5  decasteres,  and  3  centisteres. 

Ans.  2S  302005003. 

7.  4009  kilolitres  and  5  litres.  Ans.  2L  400900500. 

8.  2  hectares  and  2  centiares.  Ans.  2 A  20002. 


444  THE  METRIC   SYSTEM. 

Unit  of  Number f  Decilitre. 
9.     3002  hectolitres  and  4  millilitres.      Ans,  iL  3002000.04. 

10.  6  mjrialitres  and  1  decalitre.  Ans.  iL  600.100. 

11.  .404  millilitres.  Ans.  iL  .00004. 

DEDUCTION. 

Rule  for  Reduction  Descending.  Multiply  the  given  quantity 
by  the  number  of  the  required  denomination  which  makes  a  unit  of 
the  given  denomination. 

Since  the  multiplier  is  always  10,  100,  1000,  &c.,  the  operation 
is  performed  by  removing  the  decimal  point  as  many  places  to  the 
right  as  there  are  ci23hers  in  the  multiplier,  annexing  ciphers  when 
necessary. 

examples    for   PRACTICE. 


1.  RcMluce  ^M  32.58  to  milHmetres. 

2.  Reduce  ^M  5  to  decimetres. 

3.  Reduce  G402  to  milhgrammes. 

4.  Reduce  ^ A  42.3  to  centiares. 


5.  Reduce  "L  93.2  to  decilitres. 

6.  Reduce  *S  895  to  decasteres. 

7.  Reduce  ^A  903.2  to  mllliares. 

8.  Reduce  ^G 539  to  centii!:rammes. 


Rule  for  Reduction  Ascending.  Divide  the  given  quantity 
by  the  number  of  its  own  denomination  which  makes  a  unit  of  the 
required  denomination. 

Since  the  divisor  is  always  10,  100,  1000,  &c.,  the  operation  is 
performed  by  removing  the  decimal  point  as  many  places  to  the  left 
as  there  are  ciphers  in  the  divisor,  prefixing  ciphers  when  necessary. 


1.  Reduce  gA  5  to  myrlares. 

2.  Reduce  3M  403  to  kilometres. 

3.  Reduce  iS  42.3  to  hectosteres. 

4.  Reduce  3 A  7.2  to  decares. 


examples  for  practice. 

5.  Reduce  3G  3  to  kilogrammes. 

6.  Reduce  2L5.7  to  hectoUtres. 

7.  Reduce  3M  9  to  myriametres. 

8.  Reduce  2S47.3  to  decasteres. 


MEASURES    OF    SURFACES. 
RELATIONS  OF  UNITS  OF  SURFACE  TO    UNITS  OF  LENGTH.. 
Decimilliare  ==     One  square  decimetre  =  100  square  centimetres. 
_  f  -^^    square   decimetres,   or  a   plane  figure  whose 
~  1      length  is  one  metre  and  breadth  one  decimetre. 
Centiaro        =     One  square  miCtre  =  100  square  decimetres. 


THE   METRIC   SYSTEM. 


445 


.       _  f  10  square  metres,  or  a  plane  figure  wbose  length  is  one 
•    ^       ~  ^      decametre  and  breadth  one  metre. 
Are        =     One  square  decametre  =  100  square  metres. 
j^    ^  ^     _  (  10  square  decametres,  or  a  plane  figure  whose  length 

(is  one  hectometre  and  breadth  one  decametre. 
Hectare  =     One  square  hectometre  =  100  square  decametres. 

y^.,         (10  square  hectometres,  or  a  plane  figure  whose  length 

1      is  one  kilometre  and  breadth  one  hectometre. 
Myriare  =     One  square  kilometre  =  100  square  hectometres. 

NUMERAL  EXPRESSION    FOR  SURFACE. 

The  contents  of  a  plane  figure  is  expressed  numerically  by  giving 
the  number  of  times  it  contains  some  given  area,  which  is  assumed  as 
the  unit  of  surface. 

The  following  illustrations  will  show  how  the.  various  denomina- 
tions of  the  table  are  used  in  numerical  expressions  of  surface  :  — 

ILLUSTRATION     FIRST. 


r, 

o 

B 

1 

Length  6  metres. 

It  will  be  seen,  by  examining  this  figure,  that  the  lines  drawn 
parallel  to  the  sides,  at  the  supposed  distance  of  a  metre  from  each 
other,  divide  the  surface  into  square  metres,  and  that  there  are  as 
many  rows  of  square  metres  as  there  are  metres  in  the  breadth,  each 
row  containing  as  many  square  metres  as  there  are  metres  in  the 
length.  Hence  the  number  of  square  metres  in  the  area  of  the  figure 
is  equal  to  the  product  of  the  two  numbers  which  indicate  the  length 
and  breadth  ;  and  A  0.18  is  a  numerical  expression  for  its  contents. 


446 


THE   UETEIC   STSTEM. 


ILLUSTRATION    SECOND. 

- 

•S  o 

5  u 

'O'S^ 

''^  a 

In  this  figure,  the  lines  drawn  parallel  to  the  sides  divide  the 
figure  into  36  milliares,  or  oblongs,  whose  length  is  one  metre  and 
breadth  one  decimetre.  It  is  evident  that  ten  of  these  oblongs  put 
together  will  constitute  a  centiare,  or  square  metre.  Hence  the  ex- 
pression, 36  milliares,  may  be  written  3.6  centiares;  and  read,  three 
and  six  tenths  centiares,  or  three  centiares  and  six  milliares. 

By  reducing  the  length  to  decimetres,  the  numerical  expression  of 
the  contents  will  be,  by  Illustration  First,  60  x  6,  or  360  decimiliiares 
or  square  decimetres. 

ILLUSTRATION   THIRD. 
Length  1  decametre,  2  metres,  and  1  decimetre. 


o 

u 

a 

« 

QJ 

'O 

»-l 

n 

a 

O) 

<y 

a 

Abe. 

1 

S 

0) 

^ 

ft 

ft 

O) 

' 

B 

- 

w 

o 

a> 

B 

03 

o 

« 

-O 

1H 

.d 

'O 

a 

t 

W 

Declare. 

/■ 

Declare. 

<f 

1 1 1 1          1 i          1 j> 1 — 

Milliares.  Decimilliare. 

In  this  figure,  we  have  illustrated  the  relations  of  different  denomi- 
nations of  units  in  expressing  the  contents  of  a  given  surface^ 


THE  METRIC  SYSTEM.  417 

In  the  following  analysis,  each  part  of  the  contents  is  presented 
separately,  as  it  would  be  obtained  by  multiplying  the  length  by  the 
breadth.  The  learner  should  carefully  note  each  part,  and  analyze  a 
sufficient  number  of  examples  to  fix  the  principles  in  the  mind. 

ANALYSIS. 
Jj  ( One  decimetre  =  1  decimilliare  =  A  0.0001 

One  decimetre  x  \  Two  metres  =  2  miliiares  =  A  0.002 

(  One  decametre  =  10  miliiares  =  1  ceatiare  =  A  0.01 
(  One  decimetre  =r  2  miliiares  =  A  0.002 

Two  metres       x  \  Two  metres  =  4  centiares  =  A  0.04 

(  One  decametre  =  2  declares  =  A  0.2 

{One  decimetre  =  10  milliare  =  1  ccntiare  =  A 0.01 
Two  metres  =  2  declares  ==  A  0.2 

One  decametre  =  1  are  or  square  metre     =  A  1. 


X  = 

1—) 


^M  1.21  X 'M  1.21  =  A  1.4641 

From  these  illustrations,  we  derive  the  following  rule  for  finding  a 
numerical  expression  for  a  given  surface  of  utiiform  length  and 
breadth  :  — 

Rule.  Reduce  the  length  and  breadth  to  the  same  denomination  ; 
find  the  product  of  the  two  dimensions  after  reduction,  and  point 
off  as  many  decimal  places  in  this  product  as  there  are  decimal 
places  in  the  two  dimensions. 

The  unit  of  the  numerical  expression  thus  found  will  be  a  decimil- 
liare when  the  unit  of  length  is  a  decimetre,  a  ccntiare  when  the 
unit  of  length  is  a  metre,  an  are  when  the  unit  of  length  is  a  deca- 
metre, a  hectare  when  the  unit  of  length  is  a  hectometre,  and  a 
myriare  when  the  unit  of  length  is  a  kilometre. 

EXAMPLES   FOR   PRACTICE. 

1.  How  many  ares  in  a  floor  M  1.25  long,  and  M  8.7  wide  ? 

Ans.  A.  10875. 

2.  How  many  centiares,  how  many  kilares,  and  how  many  hec- 
tares in  the  same  floor?  Ans.  gA  10.875. 

3.  How  many  ares  in  a  board  M  5.32  by  2M  47.  ? 

Ans.  A. 025004. 

4.  How  many  miliiares,  how  many  myriares,  and  hectares  in  the 
same  board  ? 

5.  How  many  metres  of  a  carpet  nine  decimetres  wide  will  cover 


44&  THE  METRIC   SYSTEM. 

a  floor  six  metres  long  and  five  and  four-tenths  metres  wide  ?  and 
what  would  be  the  cost  of  the  carpet,  at  $2.50  a  centiare  ? 

Ans.  M  36.     $93. 

6.  In  a  farm  consisting  of  four  fields  of  the  following  dimensions, 
how  many  hectares  ?  First  field,  length  M  342,  breadth  M  273 ; 
second  field,  length  M  634,  breadth  M  350  ;  third  field,  length 
M  450,  breadth  M  329 ;  fourth  field,  length  31 730,  breadth  M  632.7. 

Ans.  2A  92.5187. 

7.  A  pile  of  lumber  was  found  to  contain  150  boards  M  4  long 
and  iM4.  wide,  225  boards  M  6.2  long  and  gM  52.  wide,  and  642 
boards  M  5.2  long  and  2M  43  wide.  How  much  was  it  worth,  at  $42. 
per  are,  face  measure.  Ans.  $1008.38  -f- 

8.  How  many  bricks  iM2.2  X  iM  1.1  would  pave  a  side- walk 
M  842.6  long  and  M  2.2  wide?  and  what  would  be  the  whole  cost 
at  82  cents  per  centiare.  Ans.  76600  bricks.     $1520.05  -\-. 

MEASUHES   OF  VOLUMES,  OU   SOLIDS. 
RELATIONS  OF  U2^ITS  OF  VOLUMES  TO  UNITS  OF  LENGTHS. 

Millistere  =     A  cubic  decimetre  =  1000  cubic  centimetres. 

r  10  cubic  decimetres,  or  a  volume,   or  solid,  whose 
Centistere  =  ■<      length  is  one  metre,  and  breadth  and  thickness  one 


1 


decimetre. 

r  10  centisteres  =  100  cubic  decimetres,  or  a  volume 
Decistere  =  -<      whose  length  and  breadth  is  one  metre,  and  thick- 

C     ness  one  decimetre. 
^^  _  f  ^  ^^^^  metre  =10  decisteres  =  100  centisteres  = 

\      1000  millisteres  or  cubic  decimetres. 

_  (10  cubic  metres,  or  a  volume  whose  leno;th  is  one 

Decastere  ^^^  ■%  *  . 

(.      decametre,  and  breadth  and  thickness  one  metre. 

(  10  decasteres  =  100  cubic  metres,  or  a  volume  whose 
Hectostere  =  -<      length  and  breath  is  one  decametre,  and  thickness 

(     one  metre. 
Kilostere    =     A  cubic  decametre  =  1000  cubic  metres. 

C  10  kilosteres,  or  a  volume  whose  length  is  one  hecto- 
Myriastere  =  ■<      metre,  and  breadth  and  thickness  each  one  doca- 

(      metre. 


THE   METRIC   SYSTEM. 


449 


NUMERICAL  EXPRESSION  FOR  VOLUME,  OR  SOLIDITY. 

The  solidity,  or  contents,  of  a  volume  is  expressed  numerically  by 
giving  the  number  of  times  it  contains  some  given  solid  as  the  unit 
of  volume. 

The  following  illustrations  will  show  how  the  various  denominations 
of  the  table  are  used  in  numerical  expressions  of  volume. 


Millistere,  or  Cubic  Decimetre, 

10  millisteres,  placed  side  by  side,  make  a  volume  whose  length 
is  one  metre,  and  breadth  and  thickness  each  one  decimetre,  thus,  — 


10  centistere,  placed  side  by  side,  make  a  volume  whose  length 
and  breadth  is  each  one  metre,  and  thickness  one  decimetre,  thus,  — 


Decistre  =  10  Centisteres  =  100  Millisteres, 

10  decisteres,  placed  face  to  face,  make  a  cube  whose  edge  is  one 
metre,  thus,  — 


Stere  =10  Decisterea  =  100  Centisteres  =  1000  Millisteres, 

From  these  illustrations,  it  is  evident  that  the  contents  of  a  cubic 
metre  may  be  expressed  numerically,  as  S  1,  iS  10, 2S  100,  ^S  1000. 


450 


THE   METRIC   SYSTEM. 


The  following  figures  illustrate  the  use  of 
the  same  four  denominations  in  expressing 
the  contents  of  a  cubic  volume  whose  edge 
is  one  metre  and  one  decimetre.  The  sur- 
face of  one  face  of  the  volume  contains 
one  centiare,  two  milliares,  and  one  deci- 
milliare,  thus,  — 


Centiare. 


Milliure. 


F 


.^^ 


Taking  a  slab  of  the  face  one  decimetre  thick,  thus,  — 
and  we  have  one  decistere,  two 
centisteres,  and  one  millistere. 
But  the  volume  is  eleven  deci- 
metres thick  ;  therefore  we  have 
eleven  such  slabs,  or  eleven  times  one  decistere,  two  centisteres,  and 
one  millistere. 

r  11  millisteres  =  1  centistere  and  1  millistere  =  S  0.011 
=  <  22  centisteres  =  2  decisteres  and  2  centisteres  =  S  0.22 
(11  decisteres   =  1  stere  and  1  decistere  =  S  1.1 

Ml.l  X  Ml.l  X  Ml.l      =  S1.331 

From  these  illustrations,  we  derive  the  following  rule  for  finding  a 
numerical  expression  for  a  given  volume  of  uniform  length,  breadth, 
and  thickness :  — 

Rule,  deduce  the  length,  breadth,  and  thickness  to  the  same 
denomination  ;  find  the  product  of  the  three  dimensions,  after  re- 
duction, and  point  off  as  many  decimal  places  in  this  product  as 
there  are  decimal  places  in  the  three  dimensions. 

The  unit  of  the  numerical  expression  thus  found  will  be  a  millistere 
when  the  unit  of  length  is  a  decimetre,  a  stere  when  the  unit  of  length 
is  a  metre,  a  kilostere^  when  the  unit  of  length  is  a  decametre. 

EXAMPLES    FOR   PRACTICE. 

1.  How  many  steres  in  a  wall  twenty-four  metres  long,  eight  and 
five-tenth  metres  high,  and  fifty-two  centimetres  thick  ?  And  what 
would  be  the  cost  of  building  it,  at  $4.25  a  stere  ? 

Ans,  S  106.08.     Cost,  $450.84. 


THE   METRIC    SYSTEM.  451 

2.  What  would  be  the  cost  of  a  pile  of  wood  fifteen  and  seven- 
tenths  metres  long,  three  metres  high,  and  seven  and  fifty- two  hun- 
dredths metres  wide,  at  $1.50  a  store?  Ans.  §531.29. 

3.  What  would  be  the  cost  of  excavating  a  cellar  eighteen  and 
three-tenths  metres  long,  ten  and  seventy-three  hundredths  metres 
wide^  and  three  and  four-tenths  metres  deep,  at  15  cents  per  store  ? 

A71S,  $100.14+. 

4.  How  deep  must  a  box  be,  whose  surface  is  thirty-two  milliares, 
to  contain  seven  and  thirty-six  hundredths  stores?       Ans.  iM  23. 

5.  How  many  stores  in  five  sticks  of  timber  of  the  following  di- 
mensions :  First,  jM  5.2  by  jM  7.3,  and  M  13  long;  second,  2M  43. 
by  2M  65,  and  M  17.5  long;  third,  iM  5.3  by  iM  3.7,  and  M  15.42 
long;  fourth,  2M  39  by  gM  56,  and  M  14  long;  fifth,  iM  4.52  by 
iM  3.78,  and  M  15  long.  Ans.  S  18.470352. 

6.  What  must  be  the  height  of  a  load  of  wood,  M  3.2  long  and 
M  1.1  wide,  to  contain  S  4.0128.  A^is.  M  1.14. 

MEASUREMENT  OF  ANGLES. 

In  the  ordinary  or  sexagesimal  system,  a  right-angle,  which  is  used 
as  the  measure  of  all  plane  angles,  is  divided  into  90  equal  parts, 
called  degrees;  a  degree  is  divided  into  60  equal  parts,  called 
minutes ;  and  a  minute  into  60  equal  parts,  called  seconds. 

In  the  centesimal  or  French  system,  a  right-angle  is  divided  into 
100  equal  parts,  called  grades ;  a  grade  into  100  equal  parts,  called 
minutes ;  and  a  minute  into  100  equal  parts,  called  seconds. 

The  former  is  called  the  sexagesimal  system,  on  account  of  the 
occurrence  of  the  number  sixty  in  forming  the  subdivisions  of  a  de- 
gree ;  and  the  latter  centesimal,  on  account  of  the  occurrence  of  the 
number  one  hundred. 

Grades,  minutes,  and  seconds  are  usually  written  thus :  35^  42^ 
24^^ ;  read,  thirty-five  grades,  forty-two  minutes,  twenty-four  seconds. 

Since  the  scale  is  centesimal,  minutes  may  be  expressed  as  hun- 
dredths, and  seconds  as  ten-thousandths ;  hence  any  number  of  grades, 
minutes,  and  seconds  may  be  expressed  decimally  thus  :  73^  4569 ; 
read,  seventy-three  grades,  forty-five  minutes,  sixty-nine  seconds. 


452  THE   METRIC   SYSTEM. 

In  a  rlght-angie,  there  are  100  grades,  or  90  degrees ;  hence,  for 
every  10  grades  there  are  9  degrees.  Dividing  the  10  grades  into 
9  equal  parts  or  degrees,  each  part  will  contain  1-J-  grades;  therefore 
a  degree  s  equal  to  1 J  grades.  Hence,  in  any  number  of  grades 
there  are  as  many  degrees  as  1^  is  contained  times  in  the  given 
number  of  grades ;  and,  conversely,  in  any  number  of  degrees  there 
are  1^  times  as  many  grades  as  there  are  degrees.  Hence  the  fol- 
lowing rules :  — 

TO  CHAIs^GE  THE  CENTESIMAL  MEASURE  TO  THE  SEXAGESIMAL. 

Rule.  Express  the  mmutes  and  seconds  as  a  decimal  of  a 
grade/  divide  hyl^x  the  quotient  will  express  the  number  of  de- 
grees and  decimals  of  a  degree  in  the  given  number  of  grades,  min- 
utes, and  seconds, 

EXAMPLES. 

Change  the  following  quantities  from  the  centesimal  measure  to 
the  sexagesimal. 

1.  25»  34^  42^\  Ans.  22°  48'  35.208'^ 

2.  57'93\  Ans,  3ri6.932^ 

3.  83"  13^  87^\  Ans,  74°  49'  29.388''. 

4.  3G^  98^  15^^  Ans.  33°  17'  .06". 

5.  14^15^60^  Am,  12°  44' 25.44". 

6.  90^  90^  90^\  Ans.  81°  49'  5.16". 

7.  18^  50^  25^\  Ans,  16°  39'  8.1". 

TO  CHANGE  THE  SEXAGESIMAL  MEASURE  TO  THE  CENTESIMAL. 

Rule.  Reduce  the  minutes  and  seconds  to  a  decimal  of  a  de- 
gree /  multiply  the  degrees  and  decimal  of  a  degree  by  1^:  the  pro- 
duct is  the  number  of  grades,  minutes,  and  seconds  in  the  given 
number  of  degrees,  minutes,  and  seconds, 

examples. 
Change  the  following  quantities  from  the  sexagesimal  measure  to 
the  centesimal. 

1.  3G°  18'  27".  Ans.  40^  31^  16  J^\ 

2.  56' 54".  Ans.  1«  5^  37^y^ 


3. 

27°  36'  45". 

4. 

189°  15'  20". 

5. 

C3°  14'  58". 

6. 

147°  24'  48". 

7. 

117°  36'  54'. 

THE   METRIC   SYSTEM.  453 

Ans.  30«  G8^  5f . 
Ans.  210^  28^  39|  V\ 

Ans.  70^  27'  71|f\ 
Ans.  168^  79' 25|f'\ 

Ans.  130«  68'  33^''. 


TO  CHANGE  THE  METRIC  TO  THE  COMMON  SYSTEM. 

Rule.  Reduce  the  given  quantity  to  the  denomination  of  the 
principal  unit  of  the  table  ;  multiply  hy  the  equivalent ,  and  reduce 
the  product  to  the  required  denomination, 

1.  ^M  3.6,  how  many  feet? 

OPERATION.  Analysis.  —  The  metre  is 

^M  3.6  X  1000  =  M  3600  *^e  principal  unit  of  the  table ; 

39.37  in.  X  3600  ==  141732  in.       ^^"^^  we- reduce  the  kilome- 

141732  in.  --  12  in.  =  11811  ft.     *^^'  ^  "^J^'^^'  .  ^^""'^  ^^^l^ 

are  39.37  inches  in  a  metre,  in 

3600  metres  there  are  3600  times  39.37  inches;  and  since  there  are 

1 2  inches  in  a  foot,  there  are  as  many  feet  as  1 2  inches  is  contained 

times  in  141732  inches.     Therefore  ^M  3.6  is  equal  to  11811  feet. 

EXAMPLES    FOR   PRACTICE. 

2.  How  many  feet  in  472  centimetres  ?       Ans,   15.4855  J  ft. 

3.  How  many  cubic  feet  in  2  kilosteres?     Ans.  70632  cu.  ft. 

4.  How  many  gallons,  wine  measure,  in  32^5  decilitres? 

Ans.  8  gals.  2.343— qts. 

5.  How  many  gallons  in  108.24  litres  ?     Ans.  28.594  -|-  gals. 

6.  How  many  bushels  in  3262  kilolitres  ? 

Ans,  92559.25  bush. 

7.  How  many  square  yards  in  436  ares  ? 

Ans.  52145.6  sq.  yds. 

8.  In  942325  centilitres,  how  many  bushels  ? 

Ans.  267.3847  +  bush. 

9.  In  436  myriagrammes,  how  many  pounds  ? 

Ans,  9611.9314  lbs. 


454  THE   METRIC   SYSTEM. 


TO  CHANGE  FROM  THE  COMMON  TO   THE  METRIC  SYSTEM. 

Rule.  Reduce  the  given  quantity  to  the  denomination  in  which 
the  equivalent  of  the  principal  unit  of  the  metric  table  is  expressed; 
divide  by  this  equivalent,  and  reduce  the  quotient  to  the  required 
denomination. 

1.  In  10  lbs.  4  oz.  liow  many  myriagrammes  ? 

OPERATION.  Analysis.  — 

10  lbs.  4  oz.  =  10.25  lbs.  The  gramme, 

10.25  lbs.  X  7000  z=  71750  gr.  *^^    principal 

71750  gr. -^15.432  gr.  =  G  4649.43—  ""^^    ^^    *^® 

G4649.43  — --10000  =  ^G. 464943—    Ans,     *''^^^'   ''    ^^" 

pressed         in 

grains;  hence  we  reduce  the  pounds  and  ounces  to  grains.     15.432 

grains  make  one  gramme;  hence  there  are  as  many  grammes  in  71750 

grains  as  15.432  grains  is  contained  times  in  71750  grains.     And  since 

there  are  10000  grammes  in  a  myriagramme,  dividing  G  4649.43 —  by 

10000  will  give  the  myriagrammes  in  10  pounds  4  ounces.     Therefore, 

10  lbs.  4  oz.  is  equal  to  ^G  464943  — 

examples  for  practice. 

2.  In  6172.8  pounds,  how  many  decagrammes? 

Ans,  ^G  280000. 

3.  How  many  hectares  in  2392  square  yards?      Ans.  ^A  .2. 

4.  How  many  ares  in  a  square  mile  ? 

Ans.  A  25899.665552—. 

5.  How  many  millisteres  in  18924  cubic  yards? 

Ans.  sS  14467889.9082 +. 

6.  In  892  grains,  how  many  hectogrammes  ? 

^715.  2G. 578019. 

7.  In  2  miles,  6  furlongs,  39  rods,  and  5  yards,  how  many 
kilometres?  Ans.  ^M 4.626416 +. 

8.  Bought  454  bush,  wheat  at  $3,  and  sold  the  same  at  $8.75 
per  hectolitre  ;  how  many  hectolitres  did  I  sell  ?  Did  I  gain  or  lose, 
and  how  much  ? 

Ans.  2L  160.     Gain,  $38. 


THE   METRIC   SYSTEM. 


455 


MISCELLANEOUS    EXAIVIPLES. 
Required  the  footings  of  the  following  bills  :  — 

(1.) 

New  York,  April  23,  1867. 


W.  J.  Milne, 

M  122  Broadclotli, 
"  320  Bid.  Shirting, 
.   "  230  White  Flannel, 
'*  206.5  Ticking, 
•'  107.9  Blk.  Silk, 

Rec'd  Payment, 


BoH  of  L.  CooLEY  &  Son. 

@    $6.00 

.35 

.30 

.31 

*'       2^   

Ans,  $1235.975 

L.  CooLEY  &  Son. 


(2.) 


Chas.  D.  McLean, 


Buffalo,  May  1,  1867. 


Bo't  of  Wm.  Benedict. 
each    «G  30.5  @  $  2.50 


^^ 

40.00 

tt 

.32 

(t 

.38 

it 

.50 

Ans, 

$4951.00 

Wm.  Benedict. 

40  chests  Tea, 

12  sacks  Java  Coffee, 

25  bbls.  Coffee  Sugar,  each    ^G  110 

10    **     Crushed  '*        ''      ^G    95 

30  boxes  Raisins,  "       ^G    12 

JRec^d  Payment, 


3.  A  man  bought  a  lot  of  land  ^M  40  long  and  ^M  20  wide,  and 
sold  one-third  of  it.  How  many  ares  had  he  left,  and  what  was  the 
cost  of  the  lot,  at  $100  per  acre? 

Ans.  to  first,  A  53333.33J.     Ans.  to  second,  $197685.95. 

4.  A  farmer  sold  ^L  540  of  wheat  at  $6,  and  invested  the  pro- 
ceeds in  coal  at  $8  per  ton.  How  many  myriagrammes  of  coal  did 
he  purchase?  Ans.  ^G  36741.835147  +. 

5.  What  will  be  the  cost  of  a  pile  of  wood  M  42.5  long,  M  2.  high, 
M  1.9  wide,  at  $2  per  stere  ?  Ans.  $323. 


456  THE   METFJC   SYSTEM. 

6.  How  many  metres  of  shirting,  at  $.25  per  metre,  must  be 
given  in  exchange  for  ^L  300  oats,  at  $1.20  per  hectolitre? 

Ans.  M  1440. 

7.  A  grocer  buys  butter  at  $.28  per  lb.,  and  sells  it  at  $.G0  per 
kilogramme.     Does  he  gain  or  lose,  and  what  per  cent.  ? 

Ans.  Lost  2{4|-  %. 

8.  A  bin  of  wheat  measures  M  5  square,  and  M  2.5  deep.  How 
many  hectolitres  will  it  contain,  and  what  will  be  the  cost  of  the 
wheat,  at  $2  per  bushel?  A?is.  ^L  625.     $3546.875. 

9.  What  price  per  pound  is  equivalent  to  $2.50  per  ^G? 

Ans.  $11.34. 

10.  A  merchant  bought  M  240  of  silk  at  $2,  and  sold  it  at  $1.95 
per  yard.     Did  he  gain  or  lose,  and  how  much  ? 

Ans.  Gain  $31.81. 

11.  Find  the  measure  of  1^  5^^  in  decimals  of  a  degree. 

Ans.  .00945.     - 

12.  A  merchant  shipped  to  France  50  bbls.  of  coffee  sugar,  each 
containing  250  lbs.,  paying  $2  per  hundred  for  transportation.  Ho 
sold  the  sugar  at  $.34  per  kilogramme,  and  invested  the  proceeds  in 
broadcloth  at  $4  per  metre  How  many  yards  of  broadcloth  did  he 
purchase?  ^ws.  458.71 -f- yds. 

13.  The  difference  between  two  angles  is  10  grades,  and  their 
sum  is  45'**     Find  each  angle.  Ans.  18°  and  27°. 

14.  Determine  the  number  of  degrees  in  the  unit  of  angular 
measure  when  an  angle  of  66 §  grades  is  represented  by  20. 

Alls.  3^. 

15.  How  many  centiares  of  plastering  in  a  house  containing  six 
rooms  of  the  following  dimensions,  deducting  one-twelfth  for  doors, 
windows,  and  base  ?  and  what  would  be  the  cost  of  the  work  at  38 
cents  per  centiare?  First  room,  M  6.2  X  M  4.7;  second  room, 
M  4.52  X  M4  ;  third  room,  M  6  X  M  5.2 ;  fourth  room,  M  382  X 
M3.82;  fifth  room,  M7  X  M6.2;  sixth  room,  M4.5  X  M  4.25r 
Height  of  each  room,  M  3.8.    Ans.  gA  562.039  +.   $213.57  +. 


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